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+ =1 b™ b (0, 0) a h (x-h, y-k) b (h, k) a (x, y) k x E X AM P L E 1 \| Sketching the Graph of a Shifted Ellipse Sketch the graph of the ellipse and determine the coordinates of the foci. ▼ SO LUTI O N The ellipse Shifted ellipse is shifted so that its center is at 2 x 4 2 y 9 1, 2 1 2 1. It is obtained from the ellipse Ellipse with center at origin y (_1, 5) 3 (_3, 2) (_1, 2) (1, 2) 2 0 (_1, _1) x FIGURE by shifting it left 1 unit and upward 2 units. The endpoints of the minor and major axes of 0, 3 the unshifted ellipse are. We apply the required shifts to these points to obtain the corresponding points on the shifted ellipse: 2, 0 0, 3 2, 0 1 2, 0 0, 3 1 0, 0 2 2 2 1, 0 2 0 1, 3 2 2 0 1, 3 2 1, 2 1 1 2 3, 2 2 2 1 1, 5 2 1, 1 2 2 1 This helps us sketch the graph in Figure 2. To ﬁnd the foci of the shifted ellipse, we ﬁrst ﬁnd the foci of the ellipse with center at. So the foci 2 9 4 5,, we have c 15 2 9 2 4 and so b a c the origin. Since 0, 15 are A B. Shifting left 1 unit and upward 2 units, we get 0, 15 A 0, 15 B B A A A 0 1, 15 2 B 0 1, 15 2 1, 2 15 B 1, 2 15 A B A B Thus, the foci of the shifted ellipse are 1, 2 15 and B A 1, 2 15 B A ✎ Practice what you’ve learned: Do Exercise 7. ▲ ■ Shifted Parabolas Applying shifts to parabolas leads to the equations and graphs shown in Figure 3. SECT IO N 8.4 \| Shifted Conics 583 y (h, k) y (h, k) 0 x 0 x y 0 (h, k) x y (h, k) 0 x
(a) (x-h)™=4p(y-k) p>0 (b) (x-h)™=4p(y-k) p<0 (c) (y-k)™=4p(x-h) p>0 (d) (y-k)™=4p(x-h) p<0 FIGURE 3 Shifted parabolas E X AM P L E 2 \| Graphing a Shifted Parabola Determine the vertex, focus, and directrix and sketch the graph of the parabola. 2 4x 8y 28 x ▼ SO LUTI O N We complete the square in x to put this equation into one of the forms in Figure 3. x2 4x 4 8y 28 4 Add 4 to complete the square x 2 x 2 1 1 2 2 2 8y 24 2 8 y 3 1 2 This parabola opens upward with vertex at Shifted parabola 2, 3 1 2. It is obtained from the parabola 2 8y x Parabola with vertex at origin by shifting right 2 units and upward 3 units. Since 4p 8, we have p 2, so the focus is 2 units above the vertex and the directrix is 2 units below the vertex. Thus, the focus is and the directrix is y 1. The graph is shown in Figure 4. 2, 5 1 ✎ Practice what you’ve learned: Do Exercises 9 and 23. ▲ 2 y 0 F(2, 5) (2, 3) y=1 x FIGURE 4 x2 4x 8y 28 ■ Shifted Hyperbolas Applying shifts to hyperbolas leads to the equations and graphs shown in Figure 5. y 0 (h, k) y (h, k) x 0 x FIGURE 5 Shifted hyperbolas (a) (x-h)™ a™ - (y-k)™ b™ =1 ( -(b) x-h)™ b™ + ( y-k)™ a™ =1 E X AM P L E 3 \| Graphing a Shifted Hyperbola A shifted conic has the equation 9x 2 72x 16y 2 32y 16 584 CHAPTER 8 \| Conic Sections (a) Complete the square in x and y to show that the equation represents a hyperbola. (b) Find the center, vertices
, foci, and asymptotes of the hyperbola, and sketch its graph. (c) Draw the graph using a graphing calculator. ▼ SO LUTI O N (a) We complete the squares in both x and y: 16 16 9 # 16 16 # 1 2 2 144 2 2y 2 2y 1 y 1 2 8x 2 8x 16 y 1 2 16 16 16 16 Group terms and factor Complete the squares Divide this by 144 Shifted hyperbola Comparing this to Figure 5(a), we see that this is the equation of a shifted hyperbola. (b) The shifted hyperbola has center 4, 1 center 1 2 and a horizontal transverse axis. 4, 1 1 2 Its graph will have the same shape as the unshifted hyperbola 2 x 16 2 y 9 1 Hyperbola with center at origin Since a 2 16 and b 2 9, we have a 4, b 3, and 116 9 5 the vertices lie 4 units to either side of the center.. Thus, the foci lie 5 units to the left and to the right of the center, and c 2a 2 b 2 foci vertices 1, 1 1 0, 1 2 and 1 2 and 9, 1 2 1 8, 1 1 y 3 4 x 2, so the asymptotes of the The asymptotes of the unshifted hyperbola are shifted hyperbola are found as follows. asymptotes and y 2 3 4 x 2 y 3 To help us sketch the hyperbola, we draw the central box; it extends 4 units left and right from the center and 3 units upward and downward from the center. We then draw the asymptotes and complete the graph of the shifted hyperbola as shown in Figure 6(a). y 0 (4, 2) F⁄(_1, _1) (0, _1) (8, _1) x F¤(9, _1) _5 5 _7 y = –1 + 0.75 x2 – 8x œ∑∑∑∑∑∑∑ 13 y = –1 – 0.75 x2 – 8x œ∑∑∑∑∑∑∑ 3 y=_ x+2 4 (b) (4, _1) (4, _4) (a) 3 y=
x-4 4 FIGURE 6 9x2 72x 16y2 32y 16 SECT IO N 8.4 \| Shifted Conics 585 (c) To draw the graph using a graphing calculator, we need to solve for y. The given equation is a quadratic equation in y, so we use the quadratic formula to solve for y. Writing the equation in the form 16y 2 32y 9x 2 72x 16 0 we get y 32 2322 4 9x 16 2 1 1 16 2 2 1 2 4608x 32 2576x 32 32 242x 2 8x 32 1 3 4 2x 2 8x 2 72x 16 2 Quadratic Formula Expand Factor 576 from under the radical Simplify To obtain the graph of the hyperbola, we graph the functions y 1 0.75 2x 2 8x and y 1 0.75 2x 2 8x as shown in Figure 6(b). ✎ Practice what you’ve learned: Do Exercises 13 and 25. ▲ ■ The General Equation of a Shifted Conic If we expand and simplify the equations of any of the shifted conics illustrated in Figures 1, 3, and 5, then we will always obtain an equation of the form Ax 2 Cy 2 Dx Ey F 0 where A and C are not both 0. Conversely, if we begin with an equation of this form, then we can complete the square in x and y to see which type of conic section the equation represents. In some cases the graph of the equation turns out to be just a pair of lines, a single point, or there may be no graph at all. These cases are called degenerate conics. If the equation is not degenerate, then we can tell whether it represents a parabola, an ellipse, or a hyperbola simply by examining the signs of A and C, as described in the box on the next page. Kepler made the momentous discovery that the orbits of the planets are elliptical. His three great laws of planetary motion are 1. The orbit of each planet is an ellipse with the sun at one focus. 2. The line segment that joins the sun to a planet sweeps out equal areas in equal time (see the ﬁgure). 3. The square of the period of revolution of a planet is proportional to the cube of the length of the major axis of its orbit. His formulation of these laws is perhaps
the most impressive deduction from empirical data in the history of science. Johannes Kepler (1571–1630) was the ﬁrst to give a correct description of the motion of the planets. The cosmology of his time postulated complicated systems of circles moving on circles to describe these motions. Kepler sought a simpler and more harmonious description. As the ofﬁcial astronomer at the imperial court in Prague, he studied the astronomical observations of the Danish. © astronomer Tycho Brahe, whose data were the most accurate available at the time. After numerous attempts to ﬁnd a theory, 586 CHAPTER 8 \| Conic Sections GENERAL EQUATION OF A SHIFTED CONIC The graph of the equation Ax 2 Cy 2 Dx Ey F 0 where A and C are not both 0, is a conic or a degenerate conic. In the nondegenerate cases the graph is 1. a parabola if A or C is 0, 2. an ellipse if A and C have the same sign (or a circle if A C), 3. a hyperbola if A and C have opposite signs. E X AM P L E 4 \| An Equation That Leads to a Degenerate Conic Sketch the graph of the equation 9x 2 y 2 18x 6y 0 ▼ SO LUTI O N Because the coefﬁcients of x 2 and y2 are of opposite sign, this equation looks as if it should represent a hyperbola (like the equation of Example 3). To see whether this is in fact the case, we complete the squares: 2 2x 2 2x 6y 2 6y Group terms and factor 9 Complete the squares Factor Divide by 9 For this to ﬁt the form of the equation of a hyperbola, we would need a nonzero constant to the right of the equal sign. In fact, further analysis shows that this is the equation of a pair of intersecting lines 3x 6 2 1 1 2 3 or y 3 1 y 3x Take square roots x 1 3 2 y 6 _2 0 x FIGURE 7 2 y 9x 2 18x 6y 0 These lines are graphed in Figure 7. ✎ Practice what you’ve learned: Do Exercise 31. ▲ Because the equation in Example 4 looked at ﬁrst glance like the equation of a hyperbola but, in fact, turned out to represent simply a pair of lines
, we refer to its graph as a degenerate hyperbola. Degenerate ellipses and parabolas can also arise when we complete the square(s) in an equation that seems to represent a conic. For example, the equation 4x 2 y 2 8x 2y 6 0 looks as if it should represent an ellipse, because the coefﬁcients of x 2 and y2 have the same sign. But completing the squares leads to which has no solution at all (since the sum of two squares cannot be negative). This equation is therefore degenerate. 8. ▼ CONCE PTS 1. Suppose we want to graph an equation in x and y. (a) If we replace x by x 3, the graph of the equation is shifted to the x 3, the graph of the equation is shifted to the by 3 units. by 3 units. If we replace x by (b) If we replace y by y 1, the graph of the equation is shifted by 1 unit. If we replace y by y 1, the graph of the equation is shifted x2 12y x 3 2. The graphs of and Label the focus, directrix, and vertex on each parabola. 2 2 1 1 2 12 y 1 by 1 unit. are given SECT IO N 8.4 \| Shifted Conics 587 y 1 0 1 x ▼ SKI LLS 5–8 ■ Find the center, foci, and vertices of the ellipse, and determine the lengths of the major and minor axes. Then sketch the graph. 5. 2 x 9 1 2 y 5 25 2 1 6. 8. 1 x 3 16 –12 ■ Find the vertex, focus, and directrix of the parabola. Then sketch the graph y2 16x 8 2 6x 12 1 2 y y 1 12. 10. 11. 9. 2 2 x 1 2B A 3. The graphs of x2 52 y2 42 1 and 1 2 x 3 52 2 1 2 y 1 42 2 1 are given. Label the vertices and foci on each ellipse. The graphs of x2 42 y2 32 1 and 1 2 x 3 42 2 1 2 y 1 32 2 1 are given. Label the vertices, foci, and asymptotes on each hyperbola. y 1 0 1 x 13–16 ■ Find the center, foci, vertices, and asymptotes of
the hyperbola. Then sketch the graph. ✎ 13 16 2 1 14. 15. y2 1 2 x 1 4 2 1 16. 1 1 x 8 2 y 1 25 17–22 ■ Find an equation for the conic whose graph is shown. 17. x y 4 18. directrix y=_12 _6 y 5 0 x _2 0 x 2 19. y 4 0 20. y 0 _3 F(8, 0) 10 x 2 x 588 CHAPTER 8 \| Conic Sections 21. y 1 0 22. asymptote y=x+1 y 4 x 0 2 6 x _4 23–34 ■ Complete the square to determine whether the equation represents an ellipse, a parabola, a hyperbola, or a degenerate conic. If the graph is an ellipse, ﬁnd the center, foci, vertices, and lengths of the major and minor axes. If it is a parabola, ﬁnd the vertex, focus, and directrix. If it is a hyperbola, ﬁnd the center, foci, vertices, and asymptotes. Then sketch the graph of the equation. If the equation has no graph, explain why. ✎ ✎ ✎ 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 2 1 x 2y 2 4 y 9x2 36x 4y2 0 x2 4y2 2x 16y 20 x2 6x 12y 9 0 4x2 25y2 24x 250y 561 0 2x2 y2 2y 1 16x2 9y2 96x 288 0 4x2 4x 8y 9 0 x 2 2x 2 16 4 y 1 x y 2 10 2 y x 3x2 4y2 6x 24y 39 0 x2 4y2 20x 40y 300 0 2 1 1 2 35–38 ■ Use a graphing device to graph the conic. 35. 36. 37. 38. 2x2 4x y 5 0 4x2 9y2 36y 0 9x2 36 y2 36x 6y x2 4y2 4x 8y 0 39. Determine what the value of F must be if the graph of the equation 4x 2 y 2 4 x 2y 1 2 F 0 is (a) an ellip
se, (b) a single point, or (c) the empty set. 40. Find an equation for the ellipse that shares a vertex and a focus with the parabola x 2 y 100 and has its other focus at the origin. 41. This exercise deals with confocal parabolas, that is, families of parabolas that have the same focus. (a) Draw graphs of the family of parabolas (b) Show that each parabola in this family has its focus at the origin. (c) Describe the effect on the graph of moving the vertex closer to the origin. ▼ APPLICATIONS 42. Path of a Cannonball A cannon ﬁres a cannonball as shown in the ﬁgure. The path of the cannonball is a parabola with vertex at the highest point of the path. If the cannonball lands 1600 ft from the cannon and the highest point it reaches is 3200 ft above the ground, ﬁnd an equation for the path of the cannonball. Place the origin at the location of the cannon. y (feet) 3200 1600 x (feet) 43. Orbit of a Satellite A satellite is in an elliptical orbit around the earth with the center of the earth at one focus. The height of the satellite above the earth varies between 140 mi and 440 mi. Assume that the earth is a sphere with radius 3960 mi. Find an equation for the path of the satellite with the origin at the center of the earth. 440 mi 140 mi ▼ DISCOVE RY • DISCUSSION • WRITI NG 44. A Family of Confocal Conics Conics that share a focus are called confocal. Consider the family of conics that have a fo0, 1 cus at 2 next page). (a) Find equations of two different ellipses that have these and a vertex at the origin (see the ﬁgure on the 1 x 2 4p y p 2 1 2, 1 1 for p 2, 3 2, 1, 2, 1, 3 2, 2. properties. (b) Find equations of two different hyperbolas that have these properties. (c) Explain why only one parabola satisﬁes these properties. Find its equation. (d) Sketch the conics you found in parts (a), (b), and (c) on the same coordinate axes (for the hyperbolas, sketch
the top branches only). (e) How are the ellipses and hyperbolas related to the parabola? CHAPTER 8 \| REVIEW ▼ P R O P E RTI LAS Geometric Deﬁnition of a Parabola (p. 552) A parabola is the set of points in the plane that are equidistant from a ﬁxed point F (the focus) and a ﬁxed line l (the directrix). Graphs of Parabolas with Vertex at the Origin (p. 554) A parabola with vertex at the origin has an equation of the form x2 4py if its axis is vertical and an equation of the form y2 4px if its axis is horizontal. x2 4py y p p>0 x p<0 y2 4px y p<0 p>0 p x Focus 0, p 1, directrix y p 2 Focus p, 0 1, directrix x p 2 Geometric Deﬁnition of an Ellipse (p. 563) An ellipse is the set of all the points in the plane for which the sum of the distances to each of two given points F1 and F2 (the foci) is a ﬁxed constant. Graphs of Ellipses with Center at the Origin (p. 564) An ellipse with center at the origin has an equation of the form x2 a2 x2 b2 if its axis is vertical (where in each case a b 0). if its axis is horizontal and an equation of the form y2 b2 y2 a2 1 1 x2 a2 y2 b2 1 y b a>b x2 b2 y2 a2 1 y a c a>b _a _c c a x _b b x _b _c _a Foci c, 0 1, c2 a2 b2 2 Foci, c2 a2 b2 0, c 2 1 CHAPTER 8 \| Review 589 y 1 0 Eccentricity of an Ellipse (p. 567) The eccentricity of an ellipse with equation x2 a2 (where a b 0) is the number x2 b2 y2 a2 1 x y2 b2 1 or e c a c 2a2 b2. The eccentricity e of any ellipse is a where
number between 0 and 1. If e is close to 0, then the ellipse is nearly circular; the closer e gets to 1, the more elongated it becomes. Geometric Deﬁnition of a Hyperbola (p. 572) A hyperbola is the set of all those points in the plane for which the absolute value of the difference of the distances to each of two given points F1 and F2 (the foci) is a ﬁxed constant. Graphs of Hyperbolas with Center at the Origin (p. 573) A hyperbola with center at the origin has an equation of the form x2 a2 x2 b2 if its axis is horizontal and an equation of the form if its axis is vertical. y2 b2 y2 a2 1 1 x2 a2 y2 b2 1 y b _b _c _a y2 a2 1 x2 b2 y c a x _b c a _a _c b x Foci c, 0 1, c2 a2 b2 2 Asymptotes: y ; x b a Foci, c2 a2 b2 0, c 2 1 Asymptotes: y ; x a b 590 CHAPTER 8 \| Conic Sections Shifted Conics (pp. 581–585) If the vertex of a parabola or the center of an ellipse or a hyperbola does not lie at the origin but rather at the point (h, k), then we refer to the curve as a shifted conic. To ﬁnd the equation of the shifted conic, we use the “unshifted” form for the appropriate curve and replace x by xh and y by yk. General Equation of a Shifted Conic (p. 586) The graph of the equation Ax2 Cy2 Dx Ey F 0 (where A and C are not both 0) is either a conic or a degenerate conic. In the nondegenerate cases the graph is: 1. A parabola if A 0 or C 0. 2. An ellipse if A and C have the same sign (or a circle if A C). 3. A hyperbola if A and C have opposite sign. To graph a conic whose equation is given in general form, complete the square in x and y to put the equation in standard form for a par
abola, an ellipse, or a hyperbola. ▼ CO N C E P T S U M MARY Section 8.1 ■ Find geometric properties of a parabola from its equation ■ Find the equation of a parabola from some of its geometric properties Section 8.2 ■ Find geometric properties of an ellipse from its equation ■ Find the equation of an ellipse from some of its geometric properties Section 8.3 ■ Find geometric properties of a hyperbola from its equation ■ Find the equation of a hyperbola from some of its geometric properties Review Exercises 1–4, 56 25, 43–44 Review Exercises 9–12, 55 26, 45 Review Exercises 17–20 27, 47 Section 8.4 ■ Find geometric properties of a shifted conic from its equation ■ Find the equation of a shifted conic from some of its geometric properties Review Exercises 5–8, 13–16, 21–24, 31–42 28–30, 47, 48–54 ▼ E X E RC I S E S 1–8 ■ Find the vertex, focus, and directrix of the parabola, and sketch the graph. y2 4x x2 8y 0 x y2 4y 2 0 2 2x 2y 4 1 2 x x 1 12 y 2x y2 0 2x2 6x 5y 10 0 2 3 x y 6. 2. 4. 8. 1. 3. 7. 5. x 2 1 2 9–16 ■ Find the center, vertices, foci, eccentricity, and the lengths of the major and minor axes of the ellipse, and sketch the graph. 9. 11. 13. 2 2 1 y x 25 9 x2 4y2 16 x 3 2 2 y 9 16 1 2 1 15. 4x2 9y2 36y 10. 12. 14. 16. 2 2 1 y x 9 49 9x2 4y2 1 x 2 2 25 2 y y 3 16 2 2 4 1 2x 1 2 1 2 2 1 x y 2 17–24 ■ Find the center, vertices, foci, and asymptotes of the hyperbola, and sketch the graph. 17. 2 x 9 2 y 16 1 19. 21. x2 2y2 16 x 4 2 2 y 16 16 1 2 18. 20. 22. 1 2 2 y 32 1 x 49 x
2 4y2 16 23. 9y2 18y x2 6x 18 24. y2 x2 6y 25–30 ■ Find an equation for the conic whose graph is shown. 25. y 26. F(2, 0) 2 0 x _12 27. 28. y 4 0 _4 F(0, 5) x 12 x y 5 0 _5 y 8 V(4, 4) 0 4 x CHAPTER 8 \| Review 591 30. y 1 53. The path of the earth around the sun is an ellipse with the sun at one focus. The ellipse has major axis 186,000,000 mi and eccentricity 0.017. Find the distance between the earth and the sun when the earth is (a) closest to the sun and (b) farthest from the sun. 4 x 0 1 2 x 186,000,000 mi 29. y 2 0 31–42 ■ Determine the type of curve represented by the equation. Find the foci and vertices (if any), and sketch the graph. y 12 2 2 y x 144 12 x2 6x 9y2 x y 2 6 3x 1 2x2 4 4x y2 2 10 1 34. 32. y 1 2 x 12 x2 y2 144 0 x y 2 8 2 y 4x x y2 16y 38. 2x2 12x y2 6y 26 0 36x2 4y2 36x 8y 31 9x2 8y2 15x 8y 27 0 x2 4y2 4x 8 36. 2 31. 33. 35. 37. 39. 40. 41. 42. 43–52 ■ Find an equation for the conic section with the given properties. 43. The parabola with focus F and directrix y 1 0, 1 1 2 44. The parabola with vertex at the origin and focus F 5, 0 1 2 45. The ellipse with center at the origin and with x-intercepts and y-intercepts 5 46. The ellipse with center major axis of length 10 C 1 0, 4 2, foci F11 0, 0 and 0, 8 F21 2 2, and 47. The hyperbola with vertices V y 1 2 x 0, 2 1 2 and asymptotes 2 and vertices 48. The hyperbola with center C V21 1, 1 2 2 49. The ell
ipse with foci and F11 V11 2, 6 2, 4 1 2, 2 2 and on the x-axis, foci F11 2, 1 and F21 2 2, 7, 2 2 1, 3 F21 2 and with one vertex 50. The parabola with vertex 51. The ellipse with vertices 1, 8 through the point 2 52. The parabola with vertex P 1 2 2 V 1, 0 1 metry and crossing the y-axis at y 2 V 1 V11 5, 5 2 7, 12 and directrix the y-axis 7, 8 and and passing V21 2 and horizontal axis of sym- 54. A ship is located 40 mi from a straight shoreline. LORAN stations A and B are located on the shoreline, 300 mi apart. From the LORAN signals, the captain determines that the ship is 80 mi closer to A than to B. Find the location of the ship. (Place A and B on the y-axis with the x-axis halfway between them. Find the x- and y-coordinates of the ship.) A 300 mi B 40 mi 55. (a) Draw graphs of the following family of ellipses for k 1, 2, 4, and 8. 2 x 16 k 2 2 y 2 k 1 (b) Prove that all the ellipses in part (a) have the same foci. 56. (a) Draw graphs of the following family of parabolas for k 1 2, 1, 2, and 4. y kx 2 (b) Find the foci of the parabolas in part (a). (c) How does the location of the focus change as k increases? ■ CHAPTER 8 \| TEST 1. Find the focus and directrix of the parabola x 2 12y, and sketch its graph. 2. Find the vertices, foci, and the lengths of the major and minor axes for the ellipse 2 y 9 2 x 16 1. Then sketch its graph. 6. y (4, 3) 2 x 1 0 1 F(4, 0) x 2 x 16 2 y 4 1. Then sketch its graph. 3. Find the vertices, foci, and asymptotes of the hyperbola 4–6 ■ Find an equation for the conic whose graph is shown. 4. 5. (_4, 2) y 1 _1 0 x
y 2 0 7–9 ■ Sketch the graph of the equation. 7. 16x 2 36y 2 96x 36y 9 0 8. 9x 2 8y 2 36x 64y 164 9. 2x y 2 8y 8 0 10. Find an equation for the hyperbola with foci 1 11. Find an equation for the parabola with focus 12. Find an equation for the ellipse with foci 1 2 0, 5 2, 4 and with asymptotes 2 and directrix the x-axis. 1 3, 4 2 and with x-intercepts 0 and 6. y 3 4 x. 13. A parabolic reﬂector for a car headlight forms a bowl shape that is 6 in. wide at its opening and 3 in. deep, as shown in the ﬁgure. How far from the vertex should the ﬁlament of the bulb be placed if it is to be located at the focus? 6 in. 3 in. 592 ● CUMUL ATIVE REVIE W TEST \| CHAPTERS 6, 7, and 8 1. Consider the following system of equations. 2 y 2 4y 2 2y 0 x x (a) Is the system linear or nonlinear? Explain. (b) Find all solutions of the system. b (c) The graph of each equation is a conic section. Name the type of conic section in each case. (d) Graph both equations on the same set of axes. (e) On your graph, shade the region that corresponds to the solution of the system of inequalities. 2 y 2 4y 2 2y 0 x x 2. Find the complete solution of each linear system, or show that no solution exists. b (a) (b) x y z 2 2x 3y z 5 3x 5y 2z 11 y z 2 x 2y 3z 3 3x 5y 8z 7 c 3. Xavier, Yolanda, and Zachary go ﬁshing. Yolanda catches as many ﬁsh as Xavier and Zachary put together. Zachary catches 2 more ﬁsh than Xavier. The total catch for all three people is 20 ﬁsh. How many did each person catch? c 4. Let a) Calculate each of the following, or explain why the calculation can’t be done. S C A B, C D, AB
, CB, BD, det, det C S, det D C B 2 1 and det 1 2 2, which matrix, C or D, has an in2 1 D 1 (b) Based on the values you calculated for det C 1 verse? Find the inverse of the invertible one. 2 5. Consider the following system of equations. 5x 3y 5 6x 4y 0 (a) Write a matrix equation of the form AX B that is equivalent to this system. (b) Find A1, the inverse of the coefﬁcient matrix. (c) Solve the matrix equation by multiplying each side by A1. (d) Now solve the system using Cramer’s Rule. Did you get the same solution as in part (b)? b 6. Find the partial fraction decomposition of the rational function 2 7. Find the focus and directrix of each parabola, and sketch its graph. x r 1 2 6y 0 x (a) (b) x 2y 2 4y 2 4x 8 x4 4x2. 593 594 CUMUL ATIVE REVIE W TEST \| Chapters 6, 7, and 8 8. Determine whether the equation represents an ellipse or a hyperbola. If it is an ellipse, ﬁnd the coordinates of its vertices and foci, and sketch its graph. If it is a hyperbola, ﬁnd the coordinates of its vertices and foci, ﬁnd the equations of its asymptotes, and sketch its graph. (a) (b) (c. Sketch the graph of each conic section, and ﬁnd the coordinates of its foci. What type of conic section does each equation represent? 2 24y (a) 9x 2 4y 2 6x y (b) x 2 8y 16 10. Find an equation for the conic whose graph is shown. y 4 1 0 F1(0, 0) 2 5 8 F¤(10, 0 CONICS IN ARCHITECTURE Many buildings employ conic sections in their design. Architects have various reasons for using these curves, ranging from structural stability to simple beauty. But how can a huge parabola, ellipse, or hyperbola be accurately constructed in concrete and steel? In this Focus we will see how the geometric properties of the conics can be used to construct
these shapes. Conics in Buildings In ancient times architecture was part of mathematics, so architects had to be mathematicians. Many of the structures they built—pyramids, temples, amphitheaters, and irrigation projects—still stand. In modern times architects employ even more sophisticated mathematical principles. The ﬁgures below show some structures that employ conic sections in their design. Roman Amphitheater in Alexandria, Egypt (circle) © Nick Wheeler/CORBIS Ceiling of Statuary Hall in the U.S. Capitol (ellipse) Courtesy of The Architect of the Capitol Roof of the Skydome in Toronto, Canada (parabola) © Stone/Getty Images Roof of Washington Dulles Airport (hyperbola and parabola) © Richard T. Nowitz /CORBIS McDonnell Planetarium, St. Louis, MO (hyperbola) Courtesy of Chamber of Commerce, St. Louis, MO Attic in La Pedrera, Barcelona, Spain (parabola) © O. Alamany & E. Vincens/CORBIS Architects have different reasons for using conics in their designs. For example, the Spanish architect Antoni Gaudi used parabolas in the attic of La Pedrera (see photo above). He reasoned that since a rope suspended between two points with an equally distributed load (like in a suspension bridge) has the shape of a parabola, an inverted parabola would provide the best support for a ﬂat roof. Constructing Conics The equations of the conics are helpful in manufacturing small objects, because a computer-controlled cutting tool can accurately trace a curve given by an equation. But in a building project, how can we construct a portion of a parabola, ellipse, or hyperbola that spans the ceiling or walls of a building? The geometric properties of the conics provide practical ways of constructing them. For example, if you were building a circular tower, you would choose a center point, then make sure that the walls of the tower were a ﬁxed 595 596 Focus on Modeling Circle P C distance from that point. Elliptical walls can be constructed using a string anchored at two points, as shown in Figure 1. To construct a parabola, we can use the apparatus shown in Figure 2. A piece of string of length a is anchored at F and A. The T-square, also of length a, slides along the straight bar L.
A pencil at P holds the string taut against the T-square. As the T-square slides to the right the pencil traces out a curve. Ellipse P F1 F2 FIGURE 1 Constructing a circle and an ellipse F A P a Parabola L FIGURE 2 Constructing a parabola From the ﬁgure we see that F, P d 1 2 d 1 F, P L, P 2 d 2 It follows that side, we get d 1 d d 1 P, A P, A 2 1 P, A 2 d 1 2 a a The string is of length a The T-square is of length a L, P 1 2 d P, A 1 2. Subtracting d P, A 1 2 from each 2 The last equation says that the distance from F to P is equal to the distance from P to the line L. Thus, the curve is a parabola with focus F and directrix L. 1 2 1 d F, P d L, P In building projects it is easier to construct a straight line than a curve. So in some buildings, such as in the Kobe Tower (see Problem 4), a curved surface is produced by using many straight lines. We can also produce a curve using straight lines, such as the parabola shown in Figure 3. FIGURE 3 Tangent lines to a parabola Each line is tangent to the parabola; that is, the line meets the parabola at exactly one at the point point and does not cross the parabola. The line tangent to the parabola a, a y x is 2 2 1 2 y 2ax a 2 You are asked to show this in Problem 6. The parabola is called the envelope of all such lines. Conics in Architecture 597 Problems 1. Conics in Architecture The photographs on page 595 show six examples of buildings that contain conic sections. Search the Internet to ﬁnd other examples of structures that employ parabolas, ellipses, or hyperbolas in their design. Find at least one example for each type of conic. 2. Constructing a Hyperbola In this problem we construct a hyperbola. The wooden bar in the ﬁgure can pivot at F1. A string that is shorter than the bar is anchored at F2 and at A, the other end of the bar. A pencil at P holds the string taut
against the bar as it moves counterclockwise around F1. (a) Show that the curve traced out by the pencil is one branch of a hyperbola with foci at F1 and F2. (b) How should the apparatus be reconﬁgured to draw the other branch of the hyperbola? A P Pivot point F1 F2 Hyperbola 3. A Parabola in a Rectangle The following method can be used to construct a parabola that ﬁts in a given rectangle. The parabola will be approximated by many short line segments. First, draw a rectangle. Divide the rectangle in half by a vertical line segment, and label the top endpoint V. Next, divide the length and width of each half rectangle into an equal number of parts to form grid lines, as shown in the ﬁgure below. Draw lines from V to the endpoints of horizontal grid line 1, and mark the points where these lines cross the vertical grid lines labeled 1. Next, draw lines from V to the endpoints of horizontal grid line 2, and mark the points where these lines cross the vertical grid lines labeled 2. Continue in this way until you have used all the horizontal grid lines. Now use line segments to connect the points you have marked to obtain an approximation to the desired parabola. Apply this procedure to draw a parabola that ﬁts into a 6 ft by 10 ft rectangle on a lawn. Hyperbolas from Straight Lines In this problem we construct hyperbolic shapes using straight lines. Punch equally spaced holes into the edges of two large plastic lids. Connect corresponding holes with strings of equal lengths as shown in the ﬁgure on the next page. Holding the strings taut, twist one lid against the other. An imaginary surface passing through the strings has hyperbolic cross sections. (An architectural example of this is the Kobe Tower in 598 Focus on Modeling Japan, shown in the photograph.) What happens to the vertices of the hyperbolic cross sections as the lids are twisted more=x2 y a2 0 Tangent line a x F1 F2 Q2 P Q1 5. Tangent Lines to a Parabola In this problem we show that the line tangent to the parabola y x 2 at the point (a) Let m be the slope of the tangent line at x a has the equation 2 m y a a, a is.
2 1 2 1 y 2ax a 2 2. 2 1 2 a, a. Show that the equation of the tangent line (b) Use the fact that the tangent line intersects the parabola at only one point to show that 2 a, a 1 2 is the only solution of the systemc) Eliminate y from the system in part (b) to get a quadratic equation in x. Show that the dis. Since the system in (b) has exactly one solution, 2 m 2a criminant of this quadratic is 2 the discriminant must equal 0. Find m. 1 (d) Substitute the value for m you found in part (c) into the equation in part (a) and simpify to get the equation of the tangent line. 6. A Cut Cylinder In this problem we prove that when a cylinder is cut by a plane, an ellipse is formed. An architectural example of this is the Tycho Brahe Planetarium in Copenhagen (see the photograph). In the ﬁgure a cylinder is cut by a plane, resulting in the red curve. Two spheres with the same radius as the cylinder slide inside the cylinder so that they just touch the plane at F1 and F2. Choose an arbitrary point P on the curve, and let Q1 and Q2 be the two points on the cylinder where a vertical line through P touches the “equator” of each sphere. (a) Show that PF1 [Hint: Use the fact that all tangents to a sphere PQ1 and PF2 PQ2. from a given point outside the sphere are of the same length.] PQ2 is the same for all points P on the curve. (b) Explain why PQ1 (c) Show that PF1 (d) Conclude that the curve is an ellipse with foci F1 and F2. PF2 is the same for all points P on the curve.1 Sequences and Summation Notation 9.2 Arithmetic Sequences 9.3 Geometric Sequences 9.4 Mathematics of Finance 9.5 Mathematical Induction 9.6 The Binomial Theorem © CHAPTER 9 Sequences and Series Can you afford your dreams? After graduation you may be thinking about a dream job, a dream house, or a dream car. But can your dream job help you to afford your other dreams? And you don’t want to
save for years before purchasing a house or car—you want them now! Our economy makes it possible to afford our dreams immediately by giving us a way of borrowing lots of money now to be repaid later. To do this fairly, lenders must be paid interest, and borrowers must pay back the money they borrowed and the interest in a regular and timely manner. The instruments that make all these ﬁnancial transactions possible are formulas —formulas for compound interest, mortgage payments, amortization, and annuities. These formulas make our economy possible because lenders, borrowers, banks, mortgage companies, and credit card companies all use the same formulas. We will see in this chapter how these formulas are used and how they are derived using properties of sequences and series (see Section 9.4). 599599 599 600 CHAPTER 9 \| Sequences and Series 9.1 Sequences and Summation Notation LEARNING OBJECTIVES After completing this section, you will be able to: ■ Find the terms of a sequence ■ Find the terms of a recursive sequence ■ Find the partial sums of a sequence ■ Use sigma notation In this chapter we study sequences. Roughly speaking, a sequence is an inﬁnite list of numbers. The numbers in the sequence are often written as a1, a2, a3,.... The dots mean that the list continues forever. A simple example is the sequence 5, 10, 15, 20, 25,... a1 a5... a4 a3 a2 Sequences arise in many real-world situations. For example, if you deposit a sum of money into an interest-bearing account, the interest that is earned each month forms a sequence. If you drop a ball and let it bounce, the height the ball reaches at each successive bounce is a sequence. We can describe the pattern of the sequence displayed above by the following formula: 5n an You may have already thought of a different way to describe the pattern—namely, “you go from one number to the next by adding 5.” This natural way of describing the sequence is expressed by the recursive formula: an an1 5 5. Try substituting n 1, 2, 3,... in each of these formulas to see how starting with a1 they produce the numbers in the sequence. In this section we see how these different ways are used to describe speciﬁc sequences. ■ Sequences A sequence is a
set of numbers written in a speciﬁc order: a1, a2, a3, a4,..., an,... The number a1 is called the ﬁrst term, a2 is the second term, and in general an is the nth term. Since for every natural number n there is a corresponding number an, we can deﬁne a sequence as a function. DEFINITION OF A SEQUENCE A sequence is a function f whose domain is the set of natural numbers. The values f are called the terms of the sequence We usually write an instead of the function notation the number n. f n 1 2 Here is a simple example of a sequence: for the value of the function at 2, 4, 6, 8, 10,... Another way to write this sequence is to use function notation: 2n 4, a a 2, a so a 1 2 6 SE CTI O N 9. 1 \| Sequences and Summation Notation 601 The dots indicate that the sequence continues indeﬁnitely. We can write a sequence in this way when it’s clear what the subsequent terms of the sequence are. This sequence consists of even numbers. To be more accurate, however, we need to specify a procedure for ﬁnding all the terms of the sequence. This can be done by giving a formula for the nth term an of the sequence. In this case, an 2n and the sequence can be written as 2, 4, 6, 8,..., 2n,... 1st term 2nd term 3rd term 4th term nth term 2n gives all the terms of the sequence. For instance, substi- Notice how the formula an tuting 1, 2, 3, and 4 for n gives the ﬁrst four terms: 2 # 1 2 a2 2 # 3 6 a4 2 # 2 4 2 # 4 8 To ﬁnd the 103rd term of this sequence, we use n 103 to get a1 a3 2 # 103 206 a103 E X AM P L E 1 \| Finding the Terms of a Sequence Find the ﬁrst ﬁve terms and the 100th term of the sequence deﬁned by each formula. (a) an (c) tn 2n 1 n n 1 (b) cn n (d)
rn 1 2 1 1 n 2n 2 ▼ SO LUTI O N To ﬁnd the ﬁrst ﬁve terms, we substitute n 1, 2, 3, 4, and 5 in the formula for the nth term. To ﬁnd the 100th term, we substitute n 100. This gives the following. nth term (a) 2n 1 (b) n 2 1 (c) (d) n n 1 1 n 2n 2 1 First ﬁve terms 100th term 1, 3, 5, 7, 9 0, 3, 8, 15, 24 16 199 9999 100 101 1 2100, 1 32 ✎ Practice what you’ve learned: Do Exercises 3, 5, 7, and 9. ▲ In Example 1(d) the presence of n in the sequence has the effect of making suc- 1 1 2 cessive terms alternately negative and positive. It is often useful to picture a sequence by sketching its graph. Since a sequence is a function whose domain is the natural numbers, we can draw its graph in the Cartesian plane. For instance, the graph of the sequence an 1 Terms are decreasing,... FIGURE 1 is shown in Figure 1. 602 CHAPTER 9 \| Sequences and Series an 1 Terms alternate in sign. Compare the graph of the sequence shown in Figure 1 to the graph of 1 n1 shown in Figure 2. The graph of every sequence consists of isolated points that are not connected. _1 FIGURE 2 Graphing calculators are useful in analyzing sequences. To work with sequences on a TI-83, we put the calculator in SEQ mode (“sequence” mode) as in Figure 3(a). If we enter of Example 1(c), we can display the terms using the the sequence command as shown in Figure 3(b). We can also graph the sequence as shown in TABLE Figure 3(c). n 1 n/ n u 1 1 2 2 Plot1 Plot2 Plot3 Min=1 = u( ) = /( +1) n 1 2 (a) 1 2 3 4 5 6 7 =1 u( ).5.66667.75.8.83333.85714.875 (b) FIGURE 3 1.5 0 15 (c) FIGURE 3 n/ u n 1 2 1 Finding patterns is an important part of mathematics. Consider a sequence
that begins 1, 4, 9, 16,... Not all sequences can be deﬁned by a formula. For example, there is no known formula for the sequence of prime numbers: 2, 3, 5, 7, 11, 13, 17, 19, 23,... Can you detect a pattern in these numbers? In other words, can you deﬁne a sequence whose ﬁrst four terms are these numbers? The answer to this question seems easy; these numbers are the squares of the numbers 1, 2, 3, 4. Thus, the sequence we are looking for n 2. However, this is not the only sequence whose ﬁrst four terms are is deﬁned by an 1, 4, 9, 16. In other words, the answer to our problem is not unique (see Exercise 80). In the next example we are interested in ﬁnding an obvious sequence whose ﬁrst few terms agree with the given ones. E X AM P L E 2 \| Finding the nth Term of a Sequence Find the nth term of a sequence whose ﬁrst several terms are given. 1 (a) 2, 3 6, 7 (b) 2, 4, 8, 16, 32,... 4, 5 8,... ▼ SO LUTI O N (a) We notice that the numerators of these fractions are the odd numbers and the denominators are the even numbers. Even numbers are of the form 2n, and odd numbers are of the form 2n 1 (an odd number differs from an even number by 1). So a sequence that has these numbers for its ﬁrst four terms is given by an 2n 1 2n Large Prime Numbers The search for large primes fascinates many people. As of this writing, the largest known prime number is 232,582,657 1 It was discovered by a team at Central Missouri State University led by professors Curtis Cooper and Steven Boone. In decimal notation this number contains 9,808,358 digits. If it were written in full, it would occupy twice as many pages as this book contains. The team was working with a large Internet group known as GIMPS (the Great Internet Mersenne Prime Search). Numbers of the form 2 p 1, where p is prime, are called Mersenne numbers and are more easily checked for primality than others. That is why the
largest known primes are of this form. SE CTI O N 9. 1 \| Sequences and Summation Notation 603 (b) These numbers are powers of 2, and they alternate in sign, so a sequence that agrees with these terms is given by an 1 1 2 n2n You should check that these formulas do indeed generate the given terms. ✎ Practice what you’ve learned: Do Exercises 25, 27, and 29. ▲ ■ Recursively Deﬁned Sequences Some sequences do not have simple deﬁning formulas like those of the preceding example. The nth term of a sequence may depend on some or all of the terms preceding it. A sequence deﬁned in this way is called recursive. Here are two examples. E X AM P L E 3 \| Finding the Terms of a Recursively Deﬁned Sequence Find the ﬁrst ﬁve terms of the sequence deﬁned recursively by a1 1 and Eratosthenes (circa 276–195 B.C.) was a renowned Greek geographer, mathematician, and astronomer. He accurately calculated the circumference of the earth by an ingenious method. He is most famous, however, for his method for ﬁnding primes, now called the sieve of Eratosthenes. The method consists of listing the integers, beginning with 2 (the ﬁrst prime), and then crossing out all the multiples of 2, which are not prime. The next number remaining on the list is 3 (the second prime), so we again cross out all multiples of it. The next remaining number is 5 (the third prime number), and we cross out all multiples of it, and so on. In this way all numbers that are not prime are crossed out, and the remaining numbers are the primes. 7 2 6 3 4 8 5 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 3 2 1 an an1 ▼ SO LUTI O N The de
ﬁning formula for this sequence is recursive. It allows us to ﬁnd the nth term an if we know the preceding term an1. Thus, we can ﬁnd the second term from the ﬁrst term, the third term from the second term, the fourth term from the third term, and so on. Since we are given the ﬁrst term a1 2 1, we can proceed as follows. 9 3 1 2 3 2 a2 a1 1 2 1 2 a3 3 a4 3 a5 3 a2 2 a3 2 a4 33 2 2 105 2 33 105 321 2 Thus, the ﬁrst ﬁve terms of this sequence are 1, 9, 33, 105, 321,... ✎ Practice what you’ve learned: Do Exercise 13. ▲ Note that to ﬁnd the 20th term of the sequence in Example 3, we must ﬁrst ﬁnd all 19 preceding terms. This is most easily done by using a graphing calculator. Figure 4(a) shows how to enter this sequence on the TI-83 calculator. From Figure 4(b) we see that the 20th term of the sequence is a20 4,649,045,865. Plot1 Plot2 Plot3 Min=1 u( )=3(u( -1)+2) u( Min)={1} u(20) 4649045865 (a) (b) FIGURE AM P L E 4 \| The Fibonacci Sequence Find the ﬁrst 11 terms of the sequence deﬁned recursively by F1 1, F2 1 and Fn Fn1 Fn2 604 CHAPTER 9 \| Sequences and Series Fibonacci (1175–1250) was born in Pisa, Italy, and educated in North Africa. He traveled widely in the Mediterranean area and learned the various methods then in use for writing numbers. On returning to Pisa in 1202, Fibonacci advocated the use of the HinduArabic decimal system, the one we use today, over the Roman numeral system that was used in Europe in his time. His most famous book, Liber Abaci, expounds on the advantages of the HinduArabic numerals. In fact, multiplication and division were so complicated using Roman numerals that a college degree was necessary to master these skills. Interestingly, in 12
99 the city of Florence outlawed the use of the decimal system for merchants and businesses, requiring numbers to be written in Roman numerals or words. One can only speculate about the reasons for this law. ▼ SO LUTI O N To ﬁnd Fn, we need to ﬁnd the two preceding terms Fn1 and Fn2. Since we are given F1 and F2, we proceed as follows. F3 F4 F5 F2 F3 F4 F1 F2 F3 1 1 2 2 1 3 3 2 5 It’s clear what is happening here. Each term is simply the sum of the two terms that precede it, so we can easily write down as many terms as we please. Here are the ﬁrst 11 terms: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89,... ✎ Practice what you’ve learned: Do Exercise 17. ▲ The sequence in Example 4 is called the Fibonacci sequence, named after the 13th century Italian mathematician who used it to solve a problem about the breeding of rabbits (see Exercise 79). The sequence also occurs in numerous other applications in nature. (See Figures 5 and 6.) In fact, so many phenomena behave like the Fibonacci sequence that one mathematical journal, the Fibonacci Quarterly, is devoted entirely to its properties. 8 5 3 2 1 1 FIGURE 5 The Fibonacci sequence in the branching of a tree 21 34 13 3 2 5 1 1 8 FIGURE 6 Fibonacci spiral Nautilus shell SE CTI O N 9. 1 \| Sequences and Summation Notation 605 ■ The Partial Sums of a Sequence In calculus we are often interested in adding the terms of a sequence. This leads to the following deﬁnition. THE PARTIAL SUMS OF A SEQUENCE For the sequence the partial sums are a1, a2, a3, a4,..., an,... S1 S2 S3 S4 Sn a1 a1 a1 a1... a1... a2 a2 a2 a3 a3 a4 a2 a3... an S1 is called the ﬁrst partial sum, S2 is the second partial sum, and so on. Sn is called the nth partial sum. The sequence S1, S2,
S3,..., Sn,... is called the sequence of partial sums. E X AM P L E 5 \| Finding the Partial Sums of a Sequence Find the ﬁrst four partial sums and the nth partial sum of the sequence given by an ▼ SO LUTI O N The terms of the sequence are, /2n. The ﬁrst four partial sums are S1 S2 S3 S4 16 1 2 3 4 7 8 15 16 Notice that in the value of each partial sum the denominator is a power of 2 and the numerator is one less than the denominator. In general, the nth partial sum is Sn 2n 1 2n 1 1 2n Partial sums of the sequence S› Sﬁ S‹ S¤ a¤ S⁄ a⁄ Terms of the sequence 1 1 2 a‹ a› 0 1 2 3 4 aﬁ 5 n FIGURE 7 Graph of the sequence an and the sequence of partial sums Sn The ﬁrst ﬁve terms of an and Sn are graphed in Figure 7. ✎ Practice what you’ve learned: Do Exercise 37. ▲ 606 CHAPTER 9 \| Sequences and Series E X AM P L E 6 \| Finding the Partial Sums of a Sequence Find the ﬁrst four partial sums and the nth partial sum of the sequence given by an 1 n 1 n 1 ▼ SO LUTI O N The ﬁrst four partial sums are S1 1 1 2 b 1 1 2 a S2 S3 S4 Do you detect a pattern here? Of course. The nth partial sum is Sn 1 1 n 1 ✎ Practice what you’ve learned: Do Exercise 39. ▲ ■ Sigma Notation Given a sequence a1, a2, a3, a4,... we can write the sum of the ﬁrst n terms using summation notation, or sigma notation. This notation derives its name from the Greek letter (capital sigma, corresponding to our S for “sum”). Sigma notation is used as follows: n a k1 ak a1 a2 a3 a4... an The left side of this expression is read “The sum of ak from k 1 to k n.” The letter k is called the index of summation, or the
summation variable, and the idea is to replace k in the expression after the sigma by the integers 1, 2, 3,..., n, and add the resulting expressions, arriving at the right side of the equation. E X AM P L E 7 \| Sigma Notation Find each sum. (a) 5 a k1 k 2 (b) 5 a j3 1 j (c) 10 a i5 i (d) 6 a i1 2 This tells us to end with k n. This tells us to add. n a k1 ak This tells us to start with k 1. ▼ SO LUTI O N 5 a k1 (a) k 2 12 22 32 42 52 55 5 (b) a j3 1 j 1 3 1 4 1 5 47 60 SE CTI O N 9. 1 \| Sequences and Summation Notation 607 sum(seq(K2,K,1,5,1)) 55 sum(seq(1/J,J,3,5, 1)) Frac 47/60 (c) (d) 10 a i5 6 a i1 i 5 6 7 8 9 10 45 2 2 2 2 2 2 2 12 FIGURE 8 We can use a graphing calculator to evaluate sums. For instance, Figure 8 shows how the TI-83 can be used to evaluate the sums in parts (a) and (b) of Example 7. ✎ Practice what you’ve learned: Do Exercises 41 and 43. ▲ E X AM P L E 8 \| Writing Sums in Sigma Notation Write each sum using sigma notation. (a) 13 23 33 43 53 63 73 13 14 15... 177 (b) ▼ SO LUTI O N (a) We can write 13 23 33 43 53 63 73 a k1 7 3 k (b) A natural way to write this sum is 13 14 15... 177 a k3 77 1k However, there is no unique way of writing a sum in sigma notation. We could also write this sum as 13 14 15... 177 a k0 74 1k 3 or 13 14 15... 177 a k1 75 1k 2 ✎ Practice what you’ve learned: Do Exercises 61 and 63. ▲ The Golden Ratio The ancient Greeks considered a line segment to be divided into the golden ratio if the ratio of the shorter part to the longer part
is the same as the ratio of the longer part to the whole segment. 1 x Thus, the segment shown is divided into the golden ratio if The golden ratio is related to the Fibonacci sequence. In fact, it can be shown using calculus* that the ratio of two successive Fibonacci numbers Fn1 Fn gets closer to the golden ratio the larger the value of n. Try ﬁnding this ratio for n 10. 1 x x 1 x This leads to a quadratic equation whose positive solution is x 1 15 1.618 2 This ratio occurs naturally in many places. For instance, psychological experiments show that the most pleasing shape of rectangle is one whose sides are in golden ratio. The ancient Greeks agreed with this and built their temples in this ratio.618 *James Stewart, Calculus, 6th ed. (Belmont, CA: Brooks/Cole, 2007), p. 722. 608 CHAPTER 9 \| Sequences and Series The following properties of sums are natural consequences of properties of the real numbers. PROPERTIES OF SUMS Let a1, a2, a3, a4,... and b1, b2, b3, b4,... be sequences. Then for every positive integer n and any real number c, the following properties hold. 1. n a k1 1 n a k1 1 n 3. a k1 2. ak bk2 ak bk2 n a k1 n a k1 ak ak n a k1 n a k1 bk bk cak c n a a k1 ak b ▼ P RO O F To prove Property 1, we write out the left side of the equation to get n a k1 1 ak bk2 b22 Because addition is commutative and associative, we can rearrange the terms on the right side to read b12 bn 2 b32 a3 a1 a2 an... 1 1 1 1 n a k1 1 ak a1 bk2... an2 Rewriting the right side using sigma notation gives Property 1. Property 2 is proved in a similar manner. To prove Property 3, we use the Distributive Property:... bn2 a2 a3 b2 b3 b1 1 1 n a k1 cak ca1 ca2 ca3... can c 1 a1 a2 a3... an2 c n a a k1 ak b
✎ ✎ 7. an 1 n 1 2 n 2 9. an 1 1 n 2 1 11. an n n n2 8. an 1 n 2 n1 1 2 n n 1 10. an 12. an 1 3 13–18 ■ Find the ﬁrst ﬁve terms of the given recursively deﬁned sequence. 2 an1 2 and a1 3 2 ✎ 13. an 14. an 15. an 16. an 1 an1 2 2an1 1 1 an1 and a1 1 and 8 a1 1 and a1 1 ✎ 17. an 18. an an1 an1 an2 an2 and an3 a1 1, a2 a1 and 2 a2 a3 1 9. ▼ CONCE PTS 1. A sequence is a function whose domain is. 2. The nth partial sum of a sequence is the sum of the ﬁrst terms of the sequence. So for the sequence the fourth partial sum is S4. an ▼ SKI LLS 3–12 ■ Find the ﬁrst four terms and the 100th term of the sequence. ✎ ✎ 3. an n 1 5. an 1 n 1 4. an 2n 3 6. an n 2 1 SE CTI O N 9. 1 \| Sequences and Summation Notation 609 19–24 ■ Use a graphing calculator to do the following. (a) Find the ﬁrst 10 terms of the sequence. (b) Graph the ﬁrst 10 terms of the sequence. 19. an 4n 3 20. an n 2 n 57. 59. 6 a k0 100 a k3 1k 4 k x 58. 60. 9 a k6 k 1 k 3 2 n a j1 1 j1x j 1 2 21. an 23. an 24. an 12 n 1 an1 an1 and a1 an2 22. an 4 2 1 n 2 1 2 and a1 1, a2 3 ✎ ✎ ✎ ✎ ✎ 25–32 ■ Find the nth term of a sequence whose ﬁrst several terms are given. 25. 2, 4, 8, 16,... 29. 27. 1, 4, 7, 10,... 9, 7 4, 5 31. 0, 2, 0, 2, 0, 2,...
25,... 16, 9 1, 3 1 1 3, 1 9, 27, 1 26. 28. 5, 25, 125, 625,... 81,... 30. 32. 3 4, 4 1, 1 6, 6 5, 5 2, 3, 1 7,... 4, 5, 1 6,... 33–36 ■ Find the ﬁrst six partial sums S1, S2, S3, S4, S5, S6 of the sequence. 33. 1, 3, 5, 7,... 35. 1 3, 1 32, 1 33, 1 34,... 34. 12, 22, 32, 42,... 36. 1, 1, 1, 1,... 37–40 ■ Find the ﬁrst four partial sums and the nth partial sum of the sequence an. 2 3n 1 1 n 2 n 1 38. 37. an an 39. an 1n 1n 1 40. an log n n 1 b a [Hint: Use a property of logarithms to write the nth term as a difference.] 41–48 ■ Find the sum. ✎ 41. ✎ 43. 4 a k1 3 a k1 k 1 k 1 1 i 2 4 1 45. 47. 8 a i1 3 5 a k1 k1 2 42. 44. 46. 48. 4 a k1 2 k 1 j 2 100 a j1 1 12 a i4 10 3 a i1 i2i 49–54 ■ Use a graphing calculator to evaluate the sum. 3k 4 2 49. 51. 53. 10 a k1 20 a j7 2 k 1 j 2 j 1 2 22 a n0 1 1 n2n 2 50. 52. 54. 100 a k1 1 15 a j5 100 a n1 1 2 1 j 1 n n 2 1 55–60 ■ Write the sum without using sigma notation. 55. 5 a k1 1k 4 56. a i0 2i 1 2i 1 61–68 ■ Write the sum using sigma notation. 61. 1 2 3 4... 100 63. 12 22 32... 102 62. 2 4 6... 20 ✎ ✎ 64. 65. 66. 1 2 ln 2 1 3 ln 3 1 4 ln 4 1 5 ln
5... 1 100 ln 100 1 1 # 2 11 12 1 2 # 3 12 22 1 3 # 4 13 32...... 1 999 # 1000 1n n2 67 100 68. 1 2x 3x 2 4x 3 5x 4... 100x 99 69. Find a formula for the nth term of the sequence 12, 2212, 3222 12, 42322212,... [Hint: Write each term as a power of 2.] 70. Deﬁne the sequence Gn 1 15 a 1 15 1 2 n 2n 1 1 15 n 2 b TABLE command on a graphing calculator to ﬁnd Use the the ﬁrst 10 terms of this sequence. Compare to the Fibonacci sequence Fn. ▼ APPLICATIONS 71. Compound Interest Julio deposits $2000 in a savings account that pays 2.4% interest per year compounded monthly. The amount in the account after n months is given by the sequence An 2000 n 1 0.024 12 b a (a) Find the ﬁrst six terms of the sequence. (b) Find the amount in the account after 3 years. 72. Compound Interest Helen deposits $100 at the end of each month into an account that pays 6% interest per year compounded monthly. The amount of interest she has accumulated after n months is given by the sequence 100 In 1.005n 1 0.005 a n b (a) Find the ﬁrst six terms of the sequence. (b) Find the interest she has accumulated after 5 years. 73. Population of a City A city was incorporated in 2004 with a population of 35,000. It is expected that the population will increase at a rate of 2% per year. The population n years after 2004 is given by the sequence Pn 35,000 1.02 1 n 2 610 CHAPTER 9 \| Sequences and Series (a) Find the ﬁrst ﬁve terms of the sequence. (b) Find the population in 2014. 74. Paying off a Debt Margarita borrows $10,000 from her uncle and agrees to repay it in monthly installments of $200. Her uncle charges 0.5% interest per month on the balance. (a) Show that her balance An in the nth month is given recur- sively by A0 10,000 and An 1.005An1 200 (b) Find
her balance after six months. 75. Fish Farming A ﬁsh farmer has 5000 catﬁsh in his pond. The number of catﬁsh increases by 8% per month, and the farmer harvests 300 catﬁsh per month. (a) Show that the catﬁsh population Pn after n months is given recursively by P0 5000 and 1.08Pn1 (b) How many ﬁsh are in the pond after 12 months? 300 Pn 76. Price of a House The median price of a house in Orange County increases by about 6% per year. In 2002 the median price was $240,000. Let Pn be the median price n years after 2002. (a) Find a formula for the sequence Pn. (b) Find the expected median price in 2010. 77. Salary Increases A newly hired salesman is promised a beginning salary of $30,000 a year with a $2000 raise every year. Let Sn be his salary in his nth year of employment. (a) Find a recursive deﬁnition of Sn. (b) Find his salary in his ﬁfth year of employment. 78. Concentration of a Solution A biologist is trying to ﬁnd the optimal salt concentration for the growth of a certain species of mollusk. She begins with a brine solution that has 4 g/L of salt and increases the concentration by 10% every day. Let C0 denote the initial concentration and Cn the concentration after n days. (a) Find a recursive deﬁnition of Cn. (b) Find the salt concentration after 8 days. 79. Fibonacci’s Rabbits Fibonacci posed the following problem: Suppose that rabbits live forever and that every month each pair produces a new pair that becomes productive at age 2 months. If we start with one newborn pair, how many pairs of rabbits will we have in the nth month? Show that the answer is Fn, where Fn is the nth term of the Fibonacci sequence. ▼ DISCOVE RY • DISCUSSION • WRITI NG 80. Different Sequences That Start the Same (a) Show that the ﬁrst four terms of the sequence an n 2 are 1, 4, 9, 16,... (b) Show that the ﬁrst four terms of the sequence an,
4, 9, 16 are also (c) Find a sequence whose ﬁrst six terms are the same as those n 2 but whose succeeding terms differ from this of an sequence. (d) Find two different sequences that begin 2, 4, 8, 16,... 81. A Recursively Deﬁned Sequence Find the ﬁrst 40 terms of the sequence deﬁned by an1 an 2 3an if an is an even number 1 if an is an odd number 11. Do the same if a1 and a1 this type of sequence. Try several other values for a1, to test your conjecture. 25. Make a conjecture about c 82. A Different Type of Recursion Find the ﬁrst 10 terms of the sequence deﬁned by with an anan1 anan2 a1 1 and a2 1 How is this recursive sequence different from the others in this section? 9.2 Arithmetic Sequences LEARNING OBJECTIVES After completing this section, you will be able to: ■ Find the terms of an arithmetic sequence ■ Find the partial sums of an arithmetic sequence In this section we study a special type of sequence, called an arithmetic sequence. ■ Arithmetic Sequences Perhaps the simplest way to generate a sequence is to start with a number a and add to it a ﬁxed constant d, over and over again. SE CTI O N 9.2 \| Arithmetic Sequences 611 DEFINITION OF AN ARITHMETIC SEQUENCE An arithmetic sequence is a sequence of the form a, a d, a 2d, a 3d, a 4d,... The number a is the ﬁrst term, and d is the common difference of the sequence. The nth term of an arithmetic sequence is given by an a n 1 1 d 2 The number d is called the common difference because any two consecutive terms of an arithmetic sequence differ by d. E X AM P L E 1 \| Arithmetic Sequences (a) If a 2 and d 3, then we have the arithmetic sequence 2, 2 3, 2 6, 2 9,... or Any two consecutive terms of this sequence differ by d 3. The nth term is an 2, 5, 8, 11b) Consider the arithmetic sequence 1 2 9, 4, 1, 6, 11,... Here the common difference is d 5. The terms of an
arithmetic sequence decrease if the common difference is negative. The nth term is. (c) The graph of the arithmetic sequence an 1 2 n 1 1 2 Notice that the points in the graph lie on a straight line with slope d 2. an n 1 9 5 is shown in Figure 1. 1 2 20 FIGURE 1 0 10 ✎ Practice what you’ve learned: Do Exercises 5, 9, and 13. ▲ An arithmetic sequence is determined completely by the ﬁrst term a and the common difference d. Thus, if we know the ﬁrst two terms of an arithmetic sequence, then we can ﬁnd a formula for the nth term, as the next example shows. E X AM P L E 2 \| Finding Terms of an Arithmetic Sequence Find the ﬁrst six terms and the 300th term of the arithmetic sequence 13, 7,... ▼ SO LUTI O N Since the ﬁrst term is 13, we have a 13. The common difference is d 7 13 6. Thus, the nth term of this sequence is an 13 6 n 1 1 2 612 CHAPTER 9 \| Sequences and Series From this we ﬁnd the ﬁrst six terms: 13, 7, 1, 5, 11, 17,... 1781. 299 13 6 The 300th term is ✎ Practice what you’ve learned: Do Exercise 27. a300 2 1 ▲ The next example shows that an arithmetic sequence is determined completely by any two of its terms. E X AM P L E 3 \| Finding Terms of an Arithmetic Sequence The 11th term of an arithmetic sequence is 52, and the 19th term is 92. Find the 1000th term. ▼ SO LUTI O N To ﬁnd the nth term of this sequence, we need to ﬁnd a and d in the formula an a n 1 1 d 2 From this formula we get a11 Since a11 52 and a19 a a 1 11 1 19 1 2 d a 10d d a 18d a19 1 92, we get the two equations: 2 52 a 10d 92 a 18d Solving this system for a and d, we get a 2 and d 5. (Verify this.) Thus, the nth term of this sequence is e 2 5 The 1000th term is ✎ Practice what you’ve learned
: Do Exercise 37. a1000 999 1 2 an 2 5 n 1 2 1 4997. ▲ MATHEMATICS IN THE MODERN WORLD Fair Division of Assets Dividing an asset fairly among a number of people is of great interest to mathematicians. Problems of this nature include dividing the national budget, disputed land, or assets in divorce cases. In 1994 Brams and Taylor found a mathematical way of dividing things fairly. Their solution has been applied to division problems in political science, legal proceedings, and other areas. To understand the problem, consider the following example. Suppose persons A and B want to divide a property fairly between them. To divide it fairly means that both A and B must be satisﬁed with the outcome of the division. Solution: A gets to divide the property into two pieces, then B gets to choose the piece he or she wants. Since both A and B had a part in the division process, each should be satisﬁed. The situation becomes much more complicated if three or more people are involved (and that’s where mathematics comes in). Dividing things fairly involves much more than simply cutting things in half; it must take into account the relative worth each person attaches to the thing being divided. A story from the Bible illustrates this clearly. Two women appear before King Solomon, each claiming to be the mother of the same newborn baby. King Solomon’s solution is to divide the baby in half! The real mother, who attaches far more worth to the baby than anyone, immediately gives up her claim to the baby in order to save the baby’s life. Mathematical solutions to fair-division problems have recently been applied in an international treaty, the Convention on the Law of the Sea. If a country wants to develop a portion of the sea ﬂoor, it is required to divide the portion into two parts, one part to be used by itself, the other by a consortium that will preserve it for later use by a less developed country. The consortium gets ﬁrst pick. SE CTI O N 9.2 \| Arithmetic Sequences 613 ■ Partial Sums of Arithmetic Sequences Suppose we want to ﬁnd the sum of the numbers 1, 2, 3, 4,..., 100, that is, 100 a k1 k When the famous mathematician C. F. Gauss was a schoolboy, his teacher posed this problem to the class and expected that it
would keep the students busy for a long time. But Gauss answered the question almost immediately. His idea was this: Since we are adding numbers produced according to a ﬁxed pattern, there must also be a pattern (or formula) for ﬁnding the sum. He started by writing the numbers from 1 to 100 and then below them wrote the same numbers in reverse order. Writing S for the sum and adding corresponding terms gives S 1 2 3... 98 99 100 S 100 99 98... 3 2 1 2S 101 101 101... 101 101 101 It follows that 2S 100 101 10,100 and so S 5050. 1 2 Of course, the sequence of natural numbers 1, 2, 3,... is an arithmetic sequence (with a 1 and d 1), and the method for summing the ﬁrst 100 terms of this sequence can be used to ﬁnd a formula for the nth partial sum of any arithmetic sequence. We want to ﬁnd d; the sum of the ﬁrst n terms of the arithmetic sequence whose terms are that is, we want to ﬁnd a k 1 ak 1 2 Sn n a k1 Using Gauss’s method, we write 1 2 a d a 2d 1 2 a 3d Sn Sn 2Sn a a Ón 1Ôd 3 4 2a Ón 1Ôd 3 4 3 Óa dÔ a Ón 2Ôd 3 4 2a Ón 1Ôd Ón 2Ôd Óa dÔ 2a Ón 1Ôd 4 4 3 There are n identical terms on the right side of this equation, so 3 a Ón 1Ôd a 2a Ón 1Ôd 4 3 4 2Sn n 2a 3 1 n 1 d 4 2 Sn n 2 3 2a n 1 1 d 4 2 d is the nth term of this sequence. So, we can write Notice that an a n 1 1 2 Sn an 2 a b This last formula says that the sum of the ﬁrst n terms of an arithmetic sequence is the average of the ﬁrst and nth terms multiplied by n, the number of terms in the sum. We now summarize this result. PARTIAL SUMS OF AN ARITHMETIC SEQUENCE For the arithmetic sequence
the nth partial sum Sn a a d 2 1 1 a an a 2d n 1 1 d 2 a 3d is given by either of the following formulas. 1. Sn n 2 3 2a n 1 1 d 4 2 2. Sn n a an 2 a b 614 CHAPTER 9 \| Sequences and Series E X AM P L E 4 \| Finding a Partial Sum of an Arithmetic Sequence Find the sum of the ﬁrst 40 terms of the arithmetic sequence 3, 7, 11, 15,... ▼ SO LUTI O N For this arithmetic sequence, a 3 and d 4. Using Formula 1 for the partial sum of an arithmetic sequence, we get 3 2 2 1 ✎ Practice what you’ve learned: Do Exercise 43. S40 4 1 2 1 4 40 2 3 40 1 20 6 156 3240 2 ▲ E X AM P L E 5 \| Finding a Partial Sum of an Arithmetic Sequence Find the sum of the ﬁrst 50 odd numbers. ▼ SO LUTI O N The odd numbers form an arithmetic sequence with a 1 and d 2. The nth term is, so the 50th odd number is a50. Substituting in Formula 2 for the partial sum of an arithmetic se1 quence, we get 1 99 2n 1 1 2 n 1 2 50 an 1 2 2 a a50 2 ✎ Practice what you’ve learned: Do Exercise 49. 1 99 2 50 50 S50 b b a a 50 # 50 2500 ▲ E X AM P L E 6 \| Finding the Seating Capacity of an Amphitheater An amphitheater has 50 rows of seats with 30 seats in the ﬁrst row, 32 in the second, 34 in the third, and so on. Find the total number of seats. ▼ SO LUTI O N The numbers of seats in the rows form an arithmetic sequence with a 30 and d 2. Since there are 50 rows, the total number of seats is the sum S50 50 2 2 3 1 3950 30 2 49 2 1 2 4 Sn n 2 3 2a n 1 1 d 4 2 Stage Thus, the amphitheater has 3950 seats. ✎ Practice what you’ve learned: Do Exercise 65. ▲ E X AM P L E 7 \| Finding the Number of Terms in a Partial Sum How many terms of the arithmetic sequences 5, 7, 9,... must
be added to get 572? ▼ SO LUTI O N We are asked to ﬁnd n when Sn Sn 572 in Formula 1 for the partial sum of an arithmetic sequence, we get 572. Substituting a 5, d 2, and n 1 1 2 4 2 Sn n 2 3 2a n 1 1 d 4 2 2 # 5 572 n 2 572 5n n 3 1 n 1 2 0 n2 4n 572 0 n 22 n 26 1 2 1 2 This gives n 22 or n 26. But since n is the number of terms in this partial sum, we must have n 22. ✎ Practice what you’ve learned: Do Exercise 59. ▲ d is an arithmetic sequence in term is. Find the ﬁrst term. 7 2. So 38. The 12th term of an arithmetic sequence is 32, and the ﬁfth 9. ▼ CONCE PTS 1. An arithmetic sequence is a sequence in which the between successive terms is constant. ✎ 2. The sequence an a n 1 1 2 which a is the ﬁrst term and d is the for the arithmetic sequence 2 5 an n 1 1 2 the ﬁrst term is, and the common difference is. 3. True or false? The nth partial sum of an arithmetic sequence is the average of the ﬁrst and last terms times n. 4. True or false? If we know the ﬁrst and second terms of an arithmetic sequence, then we can ﬁnd all the other terms. ▼ SKI LLS 5–8 ■ A sequence is given. (a) Find the ﬁrst ﬁve terms of the sequence. (b) What is the common difference d? (c) Graph the terms you found in (a). n 1 6. 5. an an. an 2 8. an –12 ■ Find the nth term of the arithmetic sequence with given ﬁrst term a and common difference d. What is the 10th term? 9. a 3, d 5 10. a 6, d 3 ✎ 11. a 5 2, d 1 2 12. a 13, d 13 13–20 ■ Determine whether the sequence is arithmetic. If it is arithmetic, ﬁnd the common difference. ✎ 13. 5, 8, 11, 14,... 15.
2, 4, 8, 16,... 2, 0, 3 2,... 3, 3 17. 14. 3, 6, 9, 13,... 16. 2, 4, 6, 8,... 18. ln 2, ln 4, ln 8, ln 16,... 19. 2.6, 4.3, 6.0, 7.7,... 20. 1 2, 1 3, 1 4, 1 5,... 21–26 ■ Find the ﬁrst ﬁve terms of the sequence, and determine whether it is arithmetic. If it is arithmetic, ﬁnd the common difference, and express the nth term of the sequence in the standard form an a d. n 1 1 4 7n 2 21. an 23. an 25. an 1 1 2n 6n 10 22. an 24. an 26. an 4 2 n 1 n 2 1 3 1 nn 2 27–36 ■ Determine the common difference, the ﬁfth term, the nth term, and the 100th term of the arithmetic sequence. ✎ 27. 2, 5, 8, 11,... 28. 1, 5, 9, 13,... 29. 4, 9, 14, 19,... 31. 12, 8, 4, 0,... 30. 11, 8, 5, 2,... 6, 8 3, 13 3,... 6, 5 32. 7 33. 25, 26.5, 28, 29.5,... 34. 15, 12.3, 9.6, 6.9,... SE CTI O N 9.2 \| Arithmetic Sequences 615 35. 2, 2 s, 2 2s, 2 3s,... 36. t, t 3, t 6, t 9,... 37. The tenth term of an arithmetic sequence is, and the second 55 2 term is 18. Find the 20th term. 39. The 100th term of an arithmetic sequence is 98, and the com- mon difference is 2. Find the ﬁrst three terms. 40. The 20th term of an arithmetic sequence is 101, and the com- mon difference is 3. Find a formula for the nth term. 41. Which term of the arithmetic sequence
1, 4, 7,... is 88? 42. The ﬁrst term of an arithmetic sequence is 1, and the common difference is 4. Is 11,937 a term of this sequence? If so, which term is it? 43–48 ■ Find the partial sum Sn of the arithmetic sequence that satisﬁes the given conditions. 43. a 1, d 2, n 10 45. a 4, d 2, n 20 44. a 3, d 2, n 12 46. a 100, d 5, n 8 9.5, n 15 47. a1 55, d 12, n 10 8, a5 48. a2 49–54 ■ A partial sum of an arithmetic sequence is given. Find the sum. 49. 1 5 9... 401 3 2B 3... 30 0 3 2 3 A 50. 51. 0.7 2.7 4.7... 56.7 52. 10 9.9 9.8... 0.1 ✎ ✎ 53. 10 a k0 1 3 0.25k 2 54. 20 a n0 1 1 2n 2 55. Show that a right triangle whose sides are in arithmetic pro- gression is similar to a 3–4–5 triangle. 56. Find the product of the numbers 101/10, 102/10, 103/10, 104/10,..., 1019/10 57. A sequence is harmonic if the reciprocals of the terms of the sequence form an arithmetic sequence. Determine whether the following sequence is harmonic: 1, 3 7, 1 5, 3 58. The harmonic mean of two numbers is the reciprocal of the average of the reciprocals of the two numbers. Find the harmonic mean of 3 and 5. 3,... ✎ 59. An arithmetic sequence has ﬁrst term a 5 and common difference d 2. How many terms of this sequence must be added to get 2700? 60. An arithmetic sequence has ﬁrst term a1 1 and fourth term 16. How many terms of this sequence must be added to a4 get 2356? 616 CHAPTER 9 \| Sequences and Series ▼ APPLICATIONS 61. Depreciation The purchase value of an ofﬁce computer is $12,500. Its annual depreciation is $1875. Find the value of the computer after 6 years. 62. Poles in a P
ile Telephone poles are stored in a pile with 25 poles in the ﬁrst layer, 24 in the second, and so on. If there are 12 layers, how many telephone poles does the pile contain? 63. Salary Increases A man gets a job with a salary of $30,000 a year. He is promised a $2300 raise each subsequent year. Find his total earnings for a 10-year period. 64. Drive-In Theater A drive-in theater has spaces for 20 cars in the ﬁrst parking row, 22 in the second, 24 in the third, and so on. If there are 21 rows in the theater, ﬁnd the number of cars that can be parked. ✎ 65. Theater Seating An architect designs a theater with 15 seats in the ﬁrst row, 18 in the second, 21 in the third, and so on. If the theater is to have a seating capacity of 870, how many rows must the architect use in his design? 66. Falling Ball When an object is allowed to fall freely near the surface of the earth, the gravitational pull is such that the object falls 16 ft in the ﬁrst second, 48 ft in the next second, 80 ft in the next second, and so on. (a) Find the total distance a ball falls in 6 s. (b) Find a formula for the total distance a ball falls in n seconds. 67. The Twelve Days of Christmas In the well-known song “The Twelve Days of Christmas,” a person gives his sweetheart k gifts on the kth day for each of the 12 days of Christmas. The person also repeats each gift identically on each subsequent day. Thus, on the 12th day the sweetheart receives a gift for the ﬁrst day, 2 gifts for the second, 3 gifts for the third, and so on. Show that the number of gifts received on the 12th day is a partial sum of an arithmetic sequence. Find this sum. ▼ DISCOVE RY • DISCUSSION • WRITI NG 68. Arithmetic Means The arithmetic mean (or average) of two numbers a and b is m a b 2 Note that m is the same distance from a as from b, so a, m, b is an arithmetic sequence. In general, if m1, m2,..., mk are equally spaced between a and b so that a, m
1, m2,..., mk, b is an arithmetic sequence, then m1, m2,..., mk are called k arithmetic means between a and b. (a) Insert two arithmetic means between 10 and 18. (b) Insert three arithmetic means between 10 and 18. (c) Suppose a doctor needs to increase a patient’s dosage of a certain medicine from 100 mg to 300 mg per day in ﬁve equal steps. How many arithmetic means must be inserted between 100 and 300 to give the progression of daily doses, and what are these means? 9.3 Geometric Sequences LEARNING OBJECTIVES After completing this section, you will be able to: ■ Find the terms of a geometric sequence ■ Find the partial sums of a geometric sequence ■ Find the sum of an inﬁnite geometric series In this section we study geometric sequences. This type of sequence occurs frequently in applications to ﬁnance, population growth, and other ﬁelds. ■ Geometric Sequences Recall that an arithmetic sequence is generated when we repeatedly add a number d to an initial term a. A geometric sequence is generated when we start with a number a and repeatedly multiply by a ﬁxed nonzero constant r. SE CTI O N 9.3 \| Geometric Sequences 617 DEFINITION OF A GEOMETRIC SEQUENCE A geometric sequence is a sequence of the form a, ar, ar 2, ar 3, ar 4,... The number a is the ﬁrst term, and r is the common ratio of the sequence. The nth term of a geometric sequence is given by ar n1 an The number r is called the common ratio because the ratio of any two consecutive terms of the sequence is r. E X AM P L E 1 \| Geometric Sequences (a) If a 3 and r 2, then we have the geometric sequence 3, 3 # 2, 3 # 22, 3 # 23, 3 # 24,... 3, 6, 12, 24, 48,... or Notice that the ratio of any two consecutive terms is r 2. The nth term is an n1 3. (b) The sequence 2 2 1 2, 10, 50, 250, 1250,... is a geometric sequence with a 2 and r 5. When r is negative, the terms of the sequence alternate in sign. The nth
term is 2 5 n1. an 1 2 (c) The sequence 1, 1 3, 1 9 is a geometric sequence with a 1 and an (d) The graph of the geometric sequence,..., 1, 1 81 27 r 1 3 1 5. The nth term is # 2n1 1 is shown in Figure 1. Notice that 1 3B an A n1. # 2x1. y 1 5 the points in the graph lie on the graph of the exponential function If 0 r 1, then the terms of the geometric sequence ar n1 decrease, but if r 1, then the terms increase. (What happens if r 1?) ✎ Practice what you’ve learned: Do Exercises 5, 9, and 13. ▲ Geometric sequences occur naturally. Here is a simple example. Suppose a ball has elasticity such that when it is dropped, it bounces up one-third of the distance it has fallen. If this ball is dropped from a height of 2 m, then it bounces up to a height of. 3 m On its second bounce, it returns to a height of, and so on (see Figure 2). Thus, the height hn that the ball reaches on its nth bounce is given by the geometric sequence 1 3B n1 2 2 3A 2 3BA 9 m 2 2 1 3B 1 3B 1 3B hn 2 A A A n 20 0 FIGURE 1 8 h 2 m m2 3 m2 9 0 1 2 3 t FIGURE 2 lowing examples show. We can ﬁnd the nth term of a geometric sequence if we know any two terms, as the fol- E X AM P L E 2 \| Finding Terms of a Geometric Sequence Find the eighth term of the geometric sequence 5, 15, 45,.... ▼ SO LUTI O N To ﬁnd a formula for the nth term of this sequence, we need to ﬁnd a and r. Clearly, a 5. To ﬁnd r, we ﬁnd the ratio of any two consecutive terms. 618 CHAPTER 9 \| Sequences and Series For instance, r 45 15 3. Thus, 3 The eighth term is 2 ✎ Practice what you’ve learned: Do Exercise 27. a8 3 2 1 1 5 81 5 an n1 5 3 1 7 10,935. 2 ▲ E X AM P
L E 3 \| Finding Terms of a Geometric Sequence The third term of a geometric sequence is 63 4, and the sixth term is 1701 32. Find the ﬁfth term. ▼ SO LUTI O N Since this sequence is geometric, its nth term is given by the formula an ar n1. Thus, 31 ar ar ar61 ar 2 5 a3 a6 From the values we are given for these two terms, we get the following system of equations: 63 4 ar 2 u 1701 32 ar 5 Simplify We solve this system by dividing. 2 5 1701 ar 32 63 ar 4 r 3 27 8 r 3 2 Take cube root of each side ar 2 2, gives 63 Substituting for r in the ﬁrst equation, 4 a 63 4 a 7 A 3 2B It follows that the nth term of this sequence is 7 an A Solve for a n1 3 2B Thus, the ﬁfth term is 4 567 16 ✎ Practice what you’ve learned: Do Exercise 37. 7 3 2B 3 2B 51 7 a5 A A ▲ Srinivasa Ramanujan (1887–1920) was born into a poor family in the small town of Kumbakonam in India. Self-taught in mathematics, he worked in virtual isolation from other mathematicians. At the age of 25 he wrote a letter to G. H. Hardy, the leading British mathematician at the time, listing some of his discoveries. Hardy immediately recognized Ramanujan’s genius, and for the next six years the two worked together in London until Ramanujan fell ill and returned to his hometown in India, where he died a year later © Ramanujan was a genius with phenomenal ability to see hidden patterns in the properties of numbers. Most of his discoveries were written as complicated inﬁnite series, the importance of which was not recognized until many years after his death. In the last year of his life he wrote 130 pages of mysterious formulas, many of which still defy proof. Hardy tells the story that when he visited Ramanujan in a hospital and arrived in a taxi, he remarked to Ramanujan that the cab’s number, 1729, was uninteresting. Ramanujan replied “No, it is a very interesting number. It is the smallest number expressible as the sum of two cubes in two different ways.” SE CTI
O N 9.3 \| Geometric Sequences 619 ■ Partial Sums of Geometric Sequences For the geometric sequence a, ar, ar 2, ar 3, ar 4,..., ar n1,..., the nth partial sum is Sn n a k1 ar k1 a ar ar 2 ar 3 ar 4... ar n1 To ﬁnd a formula for Sn, we multiply Sn by r and subtract from Sn. n1 2 ar 2 ar 3 ar 3 ar 4... ar 4... ar n1 ar n rSn rSn Sn So Sn a ar ar ar ar a ar n 1 r Sn1 2 a Sn We summarize this result PARTIAL SUMS OF A GEOMETRIC SEQUENCE For the geometric sequence an a ar ar Sn ar n1, the nth partial sum 2 ar 4... ar 3 ar n1 r 1 1 2 is given by Sn a n 1 r 1 r E X AM P L E 4 \| Finding a Partial Sum of a Geometric Sequence Find the sum of the ﬁrst ﬁve terms of the geometric sequence 1, 0.7, 0.49, 0.343,... ▼ SO LUTI O N The required sum is the sum of the ﬁrst ﬁve terms of a geometric sequence with a 1 and r 0.7. Using the formula for Sn with n 5, we get 1 # 1 2.7731 5 S5 0.7 1 2 1 0.7 Thus, the sum of the ﬁrst ﬁve terms of this sequence is 2.7731. ✎ Practice what you’ve learned: Do Exercises 43 and 47. ▲ E X AM P L E 5 \| Finding a Partial Sum of a Geometric Sequence Find the sum 5 a k1 7 A k. 2 3B a 7 ▼ SO LUTI O N The given sum is the ﬁfth partial sum of a geometric sequence with ﬁrst. Thus, by the formula for Sn we have term # 1 32 r 2 3 1 2 3B 14 3 243 A and common ratio # 1 1 2 3B 2 3B A A 5 14 3 ✎ Practice what you’ve learned: Do Exercise 49. 14 3 S5 5 3 770 243 ▲ 620 CHAPTER 9 \| Sequences and Series ■ What
Is an Inﬁnite Series? An expression of the form a1 a2 a3 a4... is called an inﬁnite series. The dots mean that we are to continue the addition indeﬁnitely. What meaning can we attach to the sum of inﬁnitely many numbers? It seems at ﬁrst that it is not possible to add inﬁnitely many numbers and arrive at a ﬁnite number. But consider the following problem. You have a cake, and you want to eat it by ﬁrst eating half the cake, then eating half of what remains, then again eating half of what remains. This process can continue indeﬁnitely because at each stage some of the cake remains. (See Figure 3.) 16 1 8 1 4 1 2 1 16 1 32 1 8 1 4 1 2 FIGURE 3 Does this mean that it’s impossible to eat all of the cake? Of course not. Let’s write down what you have eaten from this cake: 1 16 2n... This is an inﬁnite series, and we note two things about it: First, from Figure 3 it’s clear that no matter how many terms of this series we add, the total will never exceed 1. Second, the more terms of this series we add, the closer the sum is to 1 (see Figure 3). This suggests that the number 1 can be written as the sum of inﬁnitely many smaller numbers: 1 8... 1 2n 1 1 2 1 16 1 4... To make this more precise, let’s look at the partial sums of this series S1 S2 S3 S4 16 7 8 15 16 and, in general (see Example 5 of Section 9.1), Sn 1 1 2n As n gets larger and larger, we are adding more and more of the terms of this series. Intuitively, as n gets larger, Sn gets closer to the sum of the series. Now notice that as n gets large, 1/2n gets closer and closer to 0. Thus, Sn gets close to 1 0 1. Using the notation of Section 4.6, we can write Sn 1 as n q In general, if Sn gets close to a ﬁnite number S as n gets large, we say that S is the
sum of the inﬁnite series. Here is another way to arrive at the formula for the sum of an inﬁnite geometric series: S a ar ar 2 ar 3... a ar ar 2... a r 1 a rS 2 Solve the equation S a rS for S to get S rS SE CTI O N 9.3 \| Geometric Sequences 621 ■ Inﬁnite Geometric Series An inﬁnite geometric series is a series of the form a ar ar 3 ar 2 ar 4... ar n1... We can apply the reasoning used earlier to ﬁnd the sum of an inﬁnite geometric series. The nth partial sum of such a series is given by the formula Sn It can be shown that if vince yourself of this using a calculator). It follows that Sn gets close to large, or, then r n gets close to 0 as n gets large (you can easily conas n gets 1 r a/ r 1 2 0 0 1 Sn a 1 r as n q Thus, the sum of this inﬁnite geometric series is a/ 1 r. 2 1 SUM OF AN INFINITE GEOMETRIC SERIES If r 0 0 1, then the inﬁnite geometric series 3 ar a ar ar 2 ar 4... ar n1... has the sum S a 1 r E X AM P L E 6 \| Finding the Sum of an Inﬁnite Geometric Series Find the sum of the inﬁnite geometric series 2 2 5 2 25 2 125... 2 5n... ▼ SO LUTI O N We use the formula for the sum of an inﬁnite geometric series. In this case a 2 and r 1 5. Thus, the sum of this inﬁnite series is S 2 5 2 1 1 5 ✎ Practice what you’ve learned: Do Exercise 51. ▲ MATHEMATICS IN THE MODERN WORLD Fractals Many of the things we model in this book have regular predictable shapes. But recent advances in mathematics have made it possible to model such seemingly random or even chaotic shapes as those of a cloud, a ﬂickering ﬂame, a mountain, or a jagged coastline. The basic tools in this type of modeling are the fract
als invented by the mathematician Benoit Mandelbrot. A fractal is © geometric shape built up from a simple basic shape by scaling and repeating the shape indeﬁnitely according to a given rule. Fractals have inﬁnite detail; this means the closer you look, the more you see. They are also self-similar; that is, zooming in on a portion of the fractal yields the same detail as the original shape. Because of their beautiful shapes, fractals are used by movie makers to create ﬁctional landscapes and exotic backgrounds. Although a fractal is a complex shape, it is produced according to very simple rules. This property of fractals is exploited in a process of storing pictures on a computer called fractal image compression. In this process a picture is stored as a simple basic shape and a rule; repeating the shape according to the rule produces the original picture. This is an extremely efﬁcient method of storage; that’s how thousands of color pictures can be put on a single compact disc. 622 CHAPTER 9 \| Sequences and Series E X AM P L E 7 \| Writing a Repeated Decimal as a Fraction Find the fraction that represents the rational number 2.351. ▼ SO LUTI O N This repeating decimal can be written as a series: 23 10 51 1000 51 100,000 51 10,000,000 51 1,000,000,000... After the ﬁrst term, the terms of this series form an inﬁnite geometric series with a 51 1000 and r 1 100 Thus, the sum of this part of the series is S 51 1000 1 1 100 2.351 23 10 51 1000 99 100 51 990 51 1000 2328 990 # 100 99 388 165 51 990 So ✎ Practice what you’ve learned: Do Exercise 59. 9. ▼ CONCE PTS 1. A geometric sequence is a sequence in which the successive terms is constant. of 2. The sequence an ar n1 is a geometric sequence in which a is the ﬁrst term and r is the 2 geometric sequence an 1. So for the n1 5 2 the ﬁrst term is, and the common ratio is. 3. True or false? If we know the ﬁrst and second terms of a geo- metric sequence, then we can ﬁnd all the other terms.
4. (a) The nth partial sum of a geometric sequence an ar n1 is given by r Sn 1, (b) If the sum of the inﬁnite geometric series. 0 0 a ar ar 2 ar 3 p is given by S. ▼ SKI LLS 5–8 ■ The nth term of a sequence is given. (a) Find the ﬁrst ﬁve terms of the sequence. (b) What is the common ratio r? (c) Graph the terms you found in (a). ✎ 5. 7. an an n1 5 5 2 A 2 2 1 1 2B n1 n1 6. 8. an an 4 3 1 3n1 2 ✎ ✎ 9–12 ■ Find the nth term of the geometric sequence with given ﬁrst term a and common ratio r. What is the fourth term? r 3 9. a 3, a 5 r 5 2, r a 13, r 13 10. a 6, 12. 11. 1 2 13–20 ■ Determine whether the sequence is geometric. If it is geometric, ﬁnd the common ratio. 13. 2, 4, 8, 16,... 4, 3 2, 3 8,... 3, 3 15. 14. 2, 6, 18, 36,... 16. 27, 9, 3, 1,... ▲ 18. e 2, e 4, e 6, e 8,... 6, 1 8,... 2, 1 4, 1 20. 1 1 4, 1 2, 1 3, 1 17. 5,... 19. 1.0, 1.1, 1.21, 1.331,... 21–26 ■ Find the ﬁrst ﬁve terms of the sequence, and determine whether it is geometric. If it is geometric, ﬁnd the common ratio, and express the nth term of the sequence in the standard form an an ar n1. 2 1 1 4n ln an an 21. 23. 25. 3 n 2 22. an 4 3n 5n1 2 24. an 26. an n2n 2 1 1 n n 1 27–36 ■ Determine the common ratio, the ﬁfth term, and the nth term of the geometric sequence. ✎ 27. 2
, 6, 18, 54,... 28. 29. 0.3, 0.09, 0.027, 0.0081,... 7, 14 3, 28 9, 56 27,... 1, 12, 2, 212,... 30. 31. 144, 12, 1, 1 12,... 32. 33. 3, 35/3, 37/3, 27,... 35. 1, s 2/7, s 4/ 7, s 6/7,... 1 2, 1 8,... 8, 2, 34. t 8 36. 5, 5c1, 52c1, 53c1,...,... ✎ 37. The ﬁrst term of a geometric sequence is 8, and the second term is 4. Find the ﬁfth term. 38. The ﬁrst term of a geometric sequence is 3, and the third term is. Find the ﬁfth term. 4 3 39. The common ratio in a geometric sequence is, and the fourth 2 5 term is. Find the third term. 5 2 40. The common ratio in a geometric sequence is, and the ﬁfth 3 2 term is 1. Find the ﬁrst three terms. 41. Which term of the geometric sequence 2, 6, 18,... is 118,098? 42. The second and ﬁfth terms of a geometric sequence are 10 and 1250, respectively. Is 31,250 a term of this sequence? If so, which term is it? ✎ ✎ 43–46 ■ Find the partial sum Sn of the geometric sequence that satisﬁes the given conditions. n 6 43. a 5, r 2, a 2 3, n 4 3, r 1 44. 45. a3 46. a2 28, 0.12, a6 224, n 6 a5 0.00096, n 4 47–50 ■ Find the sum. 47. 1 3 9... 2187... 1 512 1 1 2 1 8 1 4 48. ✎ 49. 10 a k0 k 3 1 2B A 50. 5 a j0 7 j 3 2B A 51–58 ■ Find the sum of the inﬁnite geometric series 27 1 9 1 27.....
. 52. 54 25 8 125... ✎ 51. 53. 55. 56.... 1 1 38 36 3 3 3 4 2 1 312 1 310 3 8 1 3 10... 57. 100 9 10 3... 58. 1 12 1 2 1 212 1 4... SE CTI O N 9.3 \| Geometric Sequences 623 69. Bouncing Ball A ball is dropped from a height of 80 ft. The elasticity of this ball is such that it rebounds three-fourths of the distance it has fallen. How high does the ball rebound on the ﬁfth bounce? Find a formula for how high the ball rebounds on the nth bounce. 70. Bacteria Culture A culture initially has 5000 bacteria, and its size increases by 8% every hour. How many bacteria are present at the end of 5 hours? Find a formula for the number of bacteria present after n hours. 71. Mixing Coolant A truck radiator holds 5 gal and is ﬁlled with water. A gallon of water is removed from the radiator and replaced with a gallon of antifreeze; then a gallon of the mixture is removed from the radiator and again replaced by a gallon of antifreeze. This process is repeated indeﬁnitely. How much water remains in the tank after this process is repeated 3 times? 5 times? n times? 72. Musical Frequencies The frequencies of musical notes (measured in cycles per second) form a geometric sequence. Middle C has a frequency of 256, and the C that is an octave higher has a frequency of 512. Find the frequency of C two octaves below middle C. 73. Bouncing Ball A ball is dropped from a height of 9 ft. The elasticity of the ball is such that it always bounces up one-third the distance it has fallen. (a) Find the total distance the ball has traveled at the instant it 59–64 ■ Express the repeating decimal as a fraction. hits the ground the ﬁfth time. ✎ 59. 0.777... 62. 2.1125 60. 0.253 63. 0.112 61. 0.030303... 64. 0.123123123... 65. If the numbers a1, a2,..., an form a geometric sequence, then a2, a3,..., an1 are geometric means between a
1 and an. Insert three geometric means between 5 and 80. 66. Find the sum of the ﬁrst ten terms of the sequence 4 4b,... 2 2b, a 3 3b, a a b, a ▼ APPLICATIONS 67. Depreciation A construction company purchases a bull- dozer for $160,000. Each year the value of the bulldozer depreciates by 20% of its value in the preceding year. Let Vn be the value of the bulldozer in the nth year. (Let n 1 be the year the bulldozer is purchased.) (a) Find a formula for Vn. (b) In what year will the value of the bulldozer be less than $100,000? 68. Family Tree A person has two parents, four grandparents, eight great-grandparents, and so on. How many ancestors does a person have 15 generations back? Father Mother Grandfather Grandmother Grandfather Grandmother (b) Find a formula for the total distance the ball has traveled at the instant it hits the ground the nth time. 74. Geometric Savings Plan A very patient woman wishes to become a billionaire. She decides to follow a simple scheme: She puts aside 1 cent the ﬁrst day, 2 cents the second day, 4 cents the third day, and so on, doubling the number of cents each day. How much money will she have at the end of 30 days? How many days will it take this woman to realize her wish? 75. St. Ives The following is a well-known children’s rhyme: As I was going to St. Ives, I met a man with seven wives; Every wife had seven sacks; Every sack had seven cats; Every cat had seven kits; Kits, cats, sacks, and wives, How many were going to St. Ives? Assuming that the entire group is actually going to St. Ives, show that the answer to the question in the rhyme is a partial sum of a geometric sequence, and ﬁnd the sum. 76. Drug Concentration A certain drug is administered once a day. The concentration of the drug in the patient’s bloodstream increases rapidly at ﬁrst, but each successive dose has less effect than the preceding one. The total amount of the drug (in mg) in the bloodstream after the nth dose is given by n a k1 50 1 2B A k1 (a
) Find the amount of the drug in the bloodstream after n 10 days. 624 CHAPTER 9 \| Sequences and Series (b) If the drug is taken on a long-term basis, the amount in the bloodstream is approximated by the inﬁnite series q k1 50 1 2B A a k1. Find the sum of this series. 77. Bouncing Ball A certain ball rebounds to half the height from which it is dropped. Use an inﬁnite geometric series to approximate the total distance the ball travels after being dropped from 1 m above the ground until it comes to rest. 78. Bouncing Ball If the ball in Exercise 77 is dropped from a height of 8 ft, then 1 s is required for its ﬁrst complete bounce—from the instant it ﬁrst touches the ground until it next touches the ground. Each subsequent complete bounce reas long as the preceding complete bounce. Use an quires inﬁnite geometric series to estimate the time interval from the instant the ball ﬁrst touches the ground until it stops bouncing. 1/ 12 79. Geometry The midpoints of the sides of a square of side 1 are joined to form a new square. This procedure is repeated for each new square. (See the ﬁgure.) (a) Find the sum of the areas of all the squares. (b) Find the sum of the perimeters of all the squares. 80. Geometry A circular disk of radius R is cut out of paper, are cut out of as shown in ﬁgure (a). Two disks of radius paper and placed on top of the ﬁrst disk, as in ﬁgure (b), and then four disks of radius ﬁgure (c). Assuming that this process can be repeated indeﬁnitely, ﬁnd the total area of all the disks. are placed on these two disks, as in 1 2 R 1 4 R (a) (b) (c) 81. Geometry A yellow square of side 1 is divided into nine smaller squares, and the middle square is colored blue as shown in the ﬁgure. Each of the smaller yellow squares is in turn divided into nine squares, and each middle square is colored blue. If this process is continued indeﬁnitely, what is the total area that is colored blue? ▼ DISCO
VE RY • DISCUSSION • WRITI NG 82. Arithmetic or Geometric? The ﬁrst four terms of a se- quence are given. Determine whether these terms can be the terms of an arithmetic sequence, a geometric sequence, or neither. Find the next term if the sequence is arithmetic or geometric. (a) 5, 3, 5, 3,... 13 13 (c), 3, 3 (e) 2, 1, 1, 2,... 2 (g) 3, 3 3, 0, 2 2 5 7 (b), 3 3 (d) 1, 1, 1, 1,... (f) x 1, x, x 1, x 2,... 16 5 (h), 1,..., 9,...,...,... 13 5 15, 1, 1 3,, 83. Reciprocals of a Geometric Sequence If a1, a2, a3,... is a geometric sequence with common ratio r, show that the sequence 1 a1, 1 a2, 1 a3,... is also a geometric sequence, and ﬁnd the common ratio. 84. Logarithms of a Geometric Sequence If a1, a2, a3,... is a geometric sequence with a common ratio r 0 and a1 show that the sequence 0, log a1, log a2, log a3,... is an arithmetic sequence, and ﬁnd the common difference. 85. Exponentials of an Arithmetic Sequence If a1, a2, a3,... is an arithmetic sequence with common difference d, show that the sequence 10a1, 10a2, 10a3,... is a geometric sequence, and ﬁnd the common ratio. DISCOVERY PR OJECT FINDING PATTERNS The ancient Greeks studied triangular numbers, square numbers, pentagonal numbers, and other polygonal numbers, like those shown in the ﬁgure. Triangular numbers 1 3 6 10 15 21 Square numbers 1 4 9 16 25 Pentagonal numbers 1 5 12 22 35 To ﬁnd a pattern for such numbers, we construct a ﬁrst difference sequence by taking differences of successive terms; we repeat the process to get a second difference sequence, third difference sequence,
and so on. For the sequence of triangular numbers Tn we get the following difference table: Triangular numbers 1 First differences 2 Second differences 3 1 3 6 1 4 10 1 5 15 1 21 6 We stop at the second difference sequence because it is a constant sequence. Assuming that this sequence will continue to have constant value 1, we can work backward from the bottom row to ﬁnd more terms of the ﬁrst difference sequence and, from these, more triangular numbers. 1. Construct a difference table for the square numbers and the pentagonal num- bers. Use your table to ﬁnd the tenth pentagonal number. 2. From the patterns you have observed so far, what do you think the second difference would be for the hexagonal numbers? Use this, together with the fact that the ﬁrst two hexagonal numbers are 1 and 6, to ﬁnd the ﬁrst eight hexagonal numbers. If a sequence is given by a polynomial function and if we calculate the ﬁrst differences, the second differences, the third differences, and so on, then eventually we get a constant sequence. 3. Construct difference tables for Cn Do the same for Fn n4. n3. Which difference sequence is constant? 4. Make up a polynomial of degree 5, and construct a difference table. Which dif- ference sequence is constant? 5. The ﬁrst few terms of a polynomial sequence are 1, 2, 4, 8, 16, 31, 57,.... Construct a difference table, and use it to ﬁnd four more terms of this sequence. 625 626 CHAPTER 9 \| Sequences and Series 9.4 Mathematics of Finance LEARNING OBJECTIVES After completing this section, you will be able to: ■ Find the amount of an annuity ■ Find the present value of an annuity ■ Find the amount of the installment payments on a loan Many ﬁnancial transactions involve payments that are made at regular intervals. For example, if you deposit $100 each month in an interest-bearing account, what will the value of your account be at the end of 5 years? If you borrow $100,000 to buy a house, how much must your monthly payments be in order to pay off the loan in 30 years? Each of these questions involves the sum of a sequence of numbers; we use the results of the preceding section
to answer them here. ■ The Amount of an Annuity An annuity is a sum of money that is paid in regular equal payments. Although the word annuity suggests annual (or yearly) payments, they can be made semiannually, quarterly, monthly, or at some other regular interval. Payments are usually made at the end of the payment interval. The amount of an annuity is the sum of all the individual payments from the time of the ﬁrst payment until the last payment is made, together with all the interest. We denote this sum by Af (the subscript f here is used to denote ﬁnal amount). E X AM P L E 1 \| Calculating the Amount of an Annuity When using interest rates in calculators, remember to convert percentages to decimals. For example, 8% is 0.08. An investor deposits $400 every December 15 and June 15 for 10 years in an account that earns interest at the rate of 8% per year, compounded semiannually. How much will be in the account immediately after the last payment? ▼ SO LUTI O N We need to ﬁnd the amount of an annuity consisting of 20 semiannual payments of $400 each. Since the interest rate is 8% per year, compounded semiannually, the interest rate per time period is i 0.08/2 0.04. The ﬁrst payment is in the account for 19 time periods, the second for 18 time periods, and so on. The last payment receives no interest. The situation can be illustrated by the time line in Figure 1. Time (years) Payment (dollars) FIGURE 1 NOW 1 2 3 … 9 10 400 400 400 400 400 400 400 400 400 400 400(1.04) 400(1.04)2 400(1.04)3 … 400(1.04)14 400(1.04)15 400(1.04)16 400(1.04)17 400(1.04)18 400(1.04)19 SE CTI O N 9.4 \| Mathematics of Finance 627 The amount Af of the annuity is the sum of these 20 amounts. Thus, 400 400 1.04 400 1.04 2... 400 1.04 1 But this is a geometric series with a 400, r 1.04, and n 20, so 2 1 1 2 Af 19 2 Af 400 20 1 1.04 1
2 1 1.04 11,911.23 Thus, the amount in the account after the last payment is $11,911.23. ✎ Practice what you’ve learned: Do Exercise 3. ▲ In general, the regular annuity payment is called the periodic rent and is denoted by R. We also let i denote the interest rate per time period and n the number of payments. We always assume that the time period in which interest is compounded is equal to the time between payments. By the same reasoning as in Example 1, we see that the amount Af of an annuity is Af n1 1 i 1 2 Since this is the nth partial sum of a geometric sequence with a R and r 1 i, the formula for the partial sum gives Af AMOUNT OF AN ANNUITY The amount Af of an annuity consisting of n regular equal payments of size R with interest rate i per time period is given by Af AM P L E 2 \| Calculating the Amount of an Annuity How much money should be invested every month at 12% per year, compounded monthly, in order to have $4000 in 18 months? ▼ SO LUTI O N In this problem i 0.12/12 0.01, Af ﬁnd the amount R of each payment. By the formula for the amount of an annuity, 4000, and n 18. We need to 4000 R 1 18 1 1 0.01 2 0.01 Solving for R, we get R 1 0.01 4000 1 1 0.01 2 18 1 2 203.928 Thus, the monthly investment should be $203.93. ✎ Practice what you’ve learned: Do Exercise 9. ▲ 628 CHAPTER 9 \| Sequences and Series ■ The Present Value of an Annuity If you were to receive $10,000 ﬁve years from now, it would be worth much less than if you got $10,000 right now. This is because of the interest you could accumulate during the next ﬁve years if you invested the money now. What smaller amount would you be willing to accept now instead of receiving $10,000 in ﬁve years? This is the amount of money that, together with interest, would be worth $10,000 in ﬁve years. The amount that we are looking for here is called the discounted value or present value. If the interest rate is 8% per year, compounded quarterly, then the
interest per time period is i 0.08/4 0.02, and there are 4 5 20 time periods. If we let PV denote the present value, then by the formula for compound interest (Section 5.1), we have so 10,000 PV PV 10,000 1 1 1 i 1 0.02 2 n PV 20 1 0.02 2 20 6729.713 1 2 Thus, in this situation the present value of $10,000 is $6729.71. This reasoning leads to is to be paid in a lump sum n time a general formula for present value. If an amount periods from now and the interest rate per time period is i, then its present value Ap is given by Af Ap Af 1 1 i n 2 Similarly, the present value of an annuity is the amount Ap that must be invested now at the interest rate i per time period to provide n payments, each of amount R. Clearly, Ap is the sum of the present values of each individual payment (see Exercise 29). Another way of ﬁnding Ap is to note that Ap is the present value of Af: Ap Af THE PRESENT VALUE OF AN ANNUITY The present value Ap of an annuity consisting of n regular equal payments of size R and interest rate i per time period is given by Ap AM P L E 3 \| Calculating the Present Value of an Annuity A person wins $10,000,000 in the California lottery, and the amount is paid in yearly installments of half a million dollars each for 20 years. What is the present value of his winnings? Assume that he can earn 10% interest, compounded annually. MATH IN THE MODERN WORLD Mathematical Economics The health of the global economy is determined by such interrelated factors as supply, demand, production, consumption, pricing, distribution, and thousands of other factors. These factors are in turn determined by economic decisions (for example, whether or not you buy a certain brand of toothpaste) made by billions of different individuals each day. How will today’s creation and distribution of goods affect tomorrow’s economy? Such questions are tackled by mathematicians who work on mathematical models of the economy. In the 1940s Wassily Leontief, a pioneer in this area, created a model consisting of thousands of equations that describe how different sectors of the economy, such as the oil industry, transportation, and communication, interact with each other. A different approach to economic models, one dealing with individuals in
the economy as opposed to large sectors, was pioneered by John Nash in the 1950s. In his model, which uses game theory, the economy is a game where individual players make decisions that often lead to mutual gain. Leontief and Nash were awarded the Nobel Prize in Economics in 1973 and 1994, respectively. Economic theory continues to be a major area of mathematical research. SE CTI O N 9.4 \| Mathematics of Finance 629 ▼ SO LUTI O N Since the amount won is paid as an annuity, we need to ﬁnd its present value. Here i 0.1, R $500,000, and n 20. Thus, Ap 500,000 1 20 1 1 0.1 0.1 2 4,256,781.859 This means that the winner really won only $4,256,781.86 if it were paid immediately. ✎ Practice what you’ve learned: Do Exercise 11. ▲ ■ Installment Buying When you buy a house or a car by installment, the payments that you make are an annuity whose present value is the amount of the loan. E X AM P L E 4 \| The Amount of a Loan A student wishes to buy a car. She can afford to pay $200 per month but has no money for a down payment. If she can make these payments for four years and the interest rate is 12%, what purchase price can she afford? ▼ SO LUTI O N The payments that the student makes constitute an annuity whose present value is the price of the car (which is also the amount of the loan, in this case). Here we have i 0.12/12 0.01, R 200, and n 12 4 48, so Ap R 1 n 1 1 i i 2 200 1 48 1 1 0.01 0.01 2 7594.792 Thus, the student can buy a car priced at $7594.79. ✎ Practice what you’ve learned: Do Exercise 19. ▲ When a bank makes a loan that is to be repaid with regular equal payments R, then the payments form an annuity whose present value Ap is the amount of the loan. So to ﬁnd the size of the payments, we solve for R in the formula for the amount of an annuity. This gives the following formula for R. INSTALLMENT BUYING If a loan Ap is to be repaid in n regular equal payments with interest rate i per time period
, then the size R of each payment is given by R iAp 1 i n 2 1 1 E X AM P L E 5 \| Calculating Monthly Mortgage Payments A couple borrows $100,000 at 9% interest as a mortage loan on a house. They expect to make monthly payments for 30 years to repay the loan. What is the size of each payment? ▼ SO LUTI O N The mortgage payments form an annuity whose present value is $100,000. Also, i 0.09/12 0.0075, and n 12 30 360. We are looking Ap for the amount R of each payment. 630 CHAPTER 9 \| Sequences and Series From the formula for installment buying, we get R iAp 1 i n 2 1 1 1 1 0.0075 100,000 2 1 1 0.0075 2 360 1 2 804.623 Thus, the monthly payments are $804.62. ✎ Practice what you’ve learned: Do Exercise 17. ▲ We now illustrate the use of graphing devices in solving problems related to installment buying. E X AM P L E 6 \| Calculating the Interest Rate from the Size of Monthly Payments A car dealer sells a new car for $18,000. He offers the buyer payments of $405 per month for 5 years. What interest rate is this car dealer charging? $18,000, ▼ SO LUTI O N The payments form an annuity with present value Ap R 405, and n 12 5 60. To ﬁnd the interest rate, we must solve for i in the equation R 1 iAp 1 i n 1 A little experimentation will convince you that it is not possible to solve this equation for i algebraically. So to ﬁnd i, we use a graphing device to graph R as a function of the interest rate x, and we then use the graph to ﬁnd the interest rate corresponding to the value of R we want ($405 in this case). Since i x/12, we graph the function 2 450 405 350 0.06 FIGURE 2 0.125 0.16 R x 1 2 x 12 1 a 2 60 18,000 1 1 x 12 b x R 350, 450 0.06, 0.16, as shown in Figure 2. We also graph the in the viewing rectangle 3 405 in the same viewing rectangle. Then, by moving the cursor to horizontal line the point of intersection of the two graphs, we
ﬁnd that the corresponding x-value is approximately 0.125. Thus, the interest rate is about %. ✎ Practice what you’ve learned: Do Exercise 25. 12 ▲ 1 2 1 2 4 3 4 9. ▼ CONCE PTS 1. An annuity is a sum of money that is paid in regular equal payments. The individual payments together with all the interest. of an annuity is the sum of all the 2. The of an annuity is the amount that must be invested now at interest rate i per time period in order to provide n payments each of amount R. ▼ APPLICATIONS 3. Annuity Find the amount of an annuity that consists of ✎ 10 annual payments of $1000 each into an account that pays 6% interest per year. 4. Annuity Find the amount of an annuity that consists of 24 monthly payments of $500 each into an account that pays 8% interest per year, compounded monthly. 5. Annuity Find the amount of an annuity that consists of 20 annual payments of $5000 each into an account that pays interest of 12% per year. 6. Annuity Find the amount of an annuity that consists of 20 semiannual payments of $500 each into an account that pays 6% interest per year, compounded semiannually. 7. Annuity Find the amount of an annuity that consists of 16 quarterly payments of $300 each into an account that pays 8% interest per year, compounded quarterly. ✎ ✎ 8. Annuity Find the amount of an annuity that consists of 40 annual payments of $2000 each into an account that pays interest of 5% per year. 9. Saving How much money should be invested every quarter at 10% per year, compounded quarterly, in order to have $5000 in 2 years? 10. Saving How much money should be invested monthly at 6% per year, compounded monthly, in order to have $2000 in 8 months? 11. Annuity What is the present value of an annuity that consists of 20 semiannual payments of $1000 at an interest rate of 9% per year, compounded semiannually? 12. Annuity What is the present value of an annuity that consists of 30 monthly payments of $300 at an interest rate of 8% per year, compounded monthly. 13. Funding an Annuity How much money must be invested now at 9% per year, compounded semiannually, to fund an annuity of 20 payments of $200
each, paid every 6 months, the ﬁrst payment being 6 months from now? 14. Funding an Annuity A 55-year-old man deposits $50,000 to fund an annuity with an insurance company. The money will be invested at 8% per year, compounded semiannually. He is to draw semiannual payments until he reaches age 65. What is the amount of each payment? 15. Financing a Car A woman wants to borrow $12,000 in order to buy a car. She wants to repay the loan by monthly installments for 4 years. If the interest rate on this loan is % per year, compounded monthly, what is the amount of each payment? 10 1 2 ✎ ✎ 16. Mortgage What is the monthly payment on a 30-year mortgage of $80,000 at 9% interest? What is the monthly payment on this same mortgage if it is to be repaid over a 15-year period? 17. Mortgage What is the monthly payment on a 30-year mortgage of $100,000 at 8% interest per year, compounded monthly? What is the total amount paid on this loan over the 30-year period? 18. Mortgage What is the monthly payment on a 15-year mortgage of $200,000 at 6% interest? What is the total amount paid on this loan over the 15-year period? 19. Mortgage Dr. Gupta is considering a 30-year mortgage at 6% interest. She can make payments of $3500 a month. What size loan can she afford? 20. Mortgage A couple can afford to make a monthly mortgage payment of $650. If the mortgage rate is 9% and the couple intends to secure a 30-year mortgage, how much can they borrow? 21. Financing a Car Jane agrees to buy a car for a down pay- ment of $2000 and payments of $220 per month for 3 years. If the interest rate is 8% per year, compounded monthly, what is the actual purchase price of her car? 22. Financing a Ring Mike buys a ring for his ﬁancee by paying $30 a month for one year. If the interest rate is 10% per year, compounded monthly, what is the price of the ring? 23. Mortgage A couple secures a 30-year loan of $100,000 at % per year, compounded monthly, to buy a house. 9 3 4 SE CTI O N 9.4 \| Mathematics of Finance 631 (
a) What is the amount of their monthly payment? (b) What total amount will they pay over the 30-year period? (c) If, instead of taking the loan, the couple deposits the monthly payments in an account that pays % interest per year, compounded monthly, how much will be in the account at the end of the 30-year period? 3 9 4 6 24. Mortgage A couple needs a mortgage of $300,000. Their mortgage broker presents them with two options: a 30-year mortgage at interest. (a) Find the monthly payment on the 30-year mortgage and on the 15-year mortgage. Which mortgage has the larger monthly payment? interest or a 15-year mortgage at 1 2% 3 4% 5 (b) Find the total amount to be paid over the life of each loan. Which mortgage has the lower total payment over its lifetime? ✎ 25. Interest Rate John buys a stereo system for $640. He agrees to pay $32 a month for 2 years. Assuming that interest is compounded monthly, what interest rate is he paying? 26. Interest Rate Janet’s payments on her $12,500 car are $420 a month for 3 years. Assuming that interest is compounded monthly, what interest rate is she paying on the car loan? 27. Interest Rate An item at a department store is priced at $189.99 and can be bought by making 20 payments of $10.50. Find the interest rate, assuming that interest is compounded monthly. 28. Interest Rate A man purchases a $2000 diamond ring for a down payment of $200 and monthly installments of $88 for 2 years. Assuming that interest is compounded monthly, what interest rate is he paying? ▼ DISCOVE RY • DISCUSSION • WRITI NG 29. Present Value of an Annuity (a) Draw a time line as in Example 1 to show that the present value of an annuity is the sum of the present values of each payment, that is, Ap b) Use part (a) to derive the formula for Ap given in the text 30. An Annuity That Lasts Forever An annuity in perpetuity is one that continues forever. Such annuities are useful in setting up scholarship funds to ensure that the award continues. (a) Draw a time line (as in Example 1) to show that to set up an annuity in perpetuity of amount R per time period, the amount that must be invested now is Ap where i
is the interest rate per time periodb) Find the sum of the inﬁnite series in part (a) to show that Ap R i (c) How much money must be invested now at 10% per year, compounded annually, to provide an annuity in perpetuity of $5000 per year? The ﬁrst payment is due in one year. 632 CHAP TER 9 \| Sequences and Series (d) How much money must be invested now at 8% per year, compounded quarterly, to provide an annuity in perpetuity of $3000 per year? The ﬁrst payment is due in one year. 31. Amortizing a Mortgage When they bought their house, John and Mary took out a $90,000 mortgage at 9% interest, repayable monthly over 30 years. Their payment is $724.17 per month (check this, using the formula in the text). The bank gave them an amortization schedule, which is a table showing how much of each payment is interest, how much goes toward the principal, and the remaining principal after each payment. The table below shows the ﬁrst few entries in the amortization schedule. Payment number Total payment Interest payment Principal Remaining payment principal 1 2 3 4 724.17 724.17 724.17 724.17 675.00 674.63 674.26 673.89 49.17 49.54 49.91 50.28 89,950.83 89,901.29 89,851.38 89,801.10 After 10 years they have made 120 payments and are wondering how much they still owe, but they have lost the amortization schedule. (a) How much do John and Mary still owe on their mortgage? [Hint: The remaining balance is the present value of the 240 remaining payments.] (b) How much of their next payment is interest and how much goes toward the principal? [Hint: Since 9% 12 0.75%, they must pay 0.75% of the remaining principal in interest each month.] 9.5 Mathematical Induction LEARNING OBJECTIVE After completing this section, you will be able to: ■ Prove a statement using the Principle of Mathematical Induction There are two aspects to mathematics—discovery and proof—and they are of equal importance. We must discover something before we can attempt to prove it, and we cannot be certain of its truth
until it has been proved. In this section we examine the relationship between these two key components of mathematics more closely. ■ Conjecture and Proof Let’s try a simple experiment. We add more and more of the odd numbers as follows 16 1 3 5 7 9 25 What do you notice about the numbers on the right side of these equations? They are, in fact, all perfect squares. These equations say the following: The sum of the ﬁrst 1 odd number is 12. The sum of the ﬁrst 2 odd numbers is 22. The sum of the ﬁrst 3 odd numbers is 32. The sum of the ﬁrst 4 odd numbers is 42. The sum of the ﬁrst 5 odd numbers is 52. Consider the polynomial Here are some values of n2 n 41 n : 2 1 43 53 71 97 41 p 47 p 61 p 83 p All the values so far are prime numbers. In fact, if you keep going, you n will ﬁnd that 2 ral numbers up to n 40. It might seem reasonable at this point to conjecis prime for every natural ture that number n. But that conjecture would be too hasty, because it is easily seen that p is not prime. This illustrates that we cannot be certain of the truth of a statement no matter how many special cases we check. We need a convincing argument—a proof—to determine the truth of a statement. is prime for all natu- 41 p p n 1 1 2 2 1 SE CTI O N 9.5 \| Mathematical Induction 633 This leads naturally to the following question: Is it true that for every natural number n, the sum of the ﬁrst n odd numbers is n 2? Could this remarkable property be true? We could try a few more numbers and ﬁnd that the pattern persists for the ﬁrst 6, 7, 8, 9, and 10 odd numbers. At this point we feel quite sure that this is always true, so we make a conjecture: The sum of the ﬁrst n odd numbers is n2. Since we know that the nth odd number is 2n 1, we can write this statement more precisely as 1 3 5... 2n 1 1 2 n2 It is important to realize that this is still a conjecture. We cannot conclude by checking a ﬁnite number of cases that
a property is true for all numbers (there are inﬁnitely many). To see this more clearly, suppose someone tells us that he has added up the ﬁrst trillion odd numbers and found that they do not add up to 1 trillion squared. What would you tell this person? It would be silly to say that you’re sure it’s true because you have already checked the ﬁrst ﬁve cases. You could, however, take out paper and pencil and start checking it yourself, but this task would probably take the rest of your life. The tragedy would be that after completing this task, you would still not be sure of the truth of the conjecture! Do you see why? Herein lies the power of mathematical proof. A proof is a clear argument that demon- strates the truth of a statement beyond doubt. ■ Mathematical Induction Let’s consider a special kind of proof called mathematical induction. Here is how it works: Suppose we have a statement that says something about all natural numbers n. Let’s call this statement P. For example, we could consider the statement P: For every natural number n, the sum of the ﬁrst n odd numbers is n2. Since this statement is about all natural numbers, it contains inﬁnitely many statements; we will call them P(1), P(2),....... : The sum of the ﬁrst 1 odd number is 12. : The sum of the ﬁrst 2 odd numbers is 22. : The sum of the ﬁrst 3 odd numbers is 32.... How can we prove all of these statements at once? Mathematical induction is a clever way of doing just that. The crux of the idea is this: Suppose we can prove that whenever one of these state- ments is true, then the one following it in the list is also true. In other words, For every k, if P is true, then P k 1 2 k 1 1 2 is true. This is called the induction step because it leads us from the truth of one statement to the next. Now, suppose that we can also prove that The induction step now leads us through the following chain of statements: is true. is true, so P is true, so P is true, so P... is true. is true. is true.... 6
34 CHAPTER 9 \| Sequences and Series So we see that if both the induction step and for all n. Here is a summary of this important method of proof. 1 P 2 1 are proved, then statement P is proved PRINCIPLE OF MATHEMATICAL INDUCTION For each natural number n, let the following two conditions are satisﬁed. P n 2 1 be a statement depending on n. Suppose that 1. P 1 is true. 1 2 2. For every natural number k, if P is true then P k 1 2 k 1 1 2 is true. Then P n 1 2 is true for all natural numbers n. To apply this principle, there are two steps: Step 1 Prove that 1 1 P Step 2 Assume that P is true. 2 k 1 2 is true, and use this assumption to prove that P k 1 1 2 is true. Notice that in Step 2 we do not prove that P k is also true. The assumption that 1 2 is true. We only show that if is true, is true is called the induction then hypothesis © We now use mathematical induction to prove that the conjecture that we made at the be- ginning of this section is true. E X AM P L E 1 \| A Proof by Mathematical Induction Prove that for all natural numbers n, 1 3 5... 2n 1 1 1 3 5... denote the statement n2 2 ▼ SO LUTI O N Let P n 1 2 Step 1 We need to show that P 1 which is of course true. 1 2 is true. But P 1 1 2 2n 1 n2. 1 2 is simply the statement that 1 12, Step 2 We assume that 1 k P 1 3 5... is true. Thus, our induction hypothesis is 2k 1 k 2 2 1 k 1 2 We want to use this to show that 2k 1 1 3 5... P 1 is true, that is by substituting k 1 for each n in the statement We start with the left side and use the induction hypothesis to obtain Note that we get 3 P the right side of the equation: n. 2 1 1 2 4 This equals k 2 by the induction hypothesis. SE CTI O N 9.5 \| Mathematical Induction 635 1 3 5... 2k 2k 2 1 2 2 3 4 4 3 2 2k 1 k 1 2 2 P Thus, 2 3 2 2k
Group the ﬁrst k terms Induction hypothesis Distributive Property Simplify Factor k 1 1 2 follows from P k 1 2, and this completes the induction step. ✎ Practice what you’ve learned: Do Exercise 3. ▲ Having proved Steps 1 and 2, we conclude by the Principle of Mathematical Induction that P n 1 2 is true for all natural numbers n. E X AM P L E 2 \| A Proof by Mathematical Induction Prove that for every natural number n ▼ SO LUTI O N Let n want to show that P 1 2 P be the statement n 1 is true for all natural numbers n2. We Step 1 We need to show that P 1 1 2 P is true. But 1 1 2 1 1 1 2 says that 1 1 2 and this statement is clearly true. k P Step 2 Assume that is true. Thus, our induction hypothesis is We want to use this to show that is true, that is This equals k 1 k 1 2 2 by the induction process. So we start with the left side and use the induction hypothesis to obtain the right side Group the ﬁrst k terms Induction hypothesis Factor k 1 b 1 2 4 Common denominator Write k 2 as k 1 1 Thus, P k 1 1 2 follows from P k 1 2 and this completes the induction step. 636 CHAPTER 9 \| Sequences and Series Having proved Steps 1 and 2, we conclude by the Principle of Mathematical Induction 1 2 n P is true for all natural numbers n. that ✎ Practice what you’ve learned: Do Exercise 5. ▲ Formulas for the sums of powers of the ﬁrst n natural numbers are important in calculus. Formula 1 in the following box is proved in Example 2. The other formulas are also proved by using mathematical induction (see Exercises 6 and 9). SUMS OF POWERS 0. 2. n a k1 n a k1 1 n n 1 n 1 k 2 2n 1 2 2 1 6 1. n a k1 k n 1 n 1 2 2 n 3. a k1 k 3 n2 2 1 n 1 4 2 It might happen that a statement from some number on. For example, we might want to prove that P Notice that if we prove that 5 would imply the truth of is false for the ﬁrst few natural numbers but true is true for n 5. is true, then this fact,
together with the induction step, 6,.... The next example illustrates this point AM P L E 3 \| Proving an Inequality by Mathematical Induction Prove that 4n 2n for all n 5. ▼ SO LUTI O N Let P n denote the statement 4n 2n. Step 1 P 5 1 Step 2 Assume that 2 1 2 is the statement that k P 4 # 5 1 2 25, or 20 32, which is true. is true. Thus, our induction hypothesis is 4k 2k We get P(k 1) by replacing k by k 1 in the statement P(k). P We want to use this to show that k 1 4 1 2 is true, that is, k 1 2 1 2k1 © Blaise Pascal (1623–1662) is considered one of the most versatile minds in modern history. He was a writer and philosopher as well as a gifted mathematician and physicist. Among his contributions that appear in this book are Pascal’s triangle and the Principle of Mathematical Induction. Pascal’s father, himself a mathematician, believed that his son should not study mathematics until he was 15 or 16. But at age 12, Blaise insisted on learning geometry and proved most of its elementary theorems himself. At 19 he invented the ﬁrst mechanical adding machine. In 1647, after writing a major treatise on the conic sections, he abruptly abandoned mathematics because he felt that his intense studies were contributing to his ill health. He devoted himself instead to frivolous recreations such as gambling, but this only served to pique his interest in probability. In 1654 he miraculously survived a carriage accident in which his horses ran off a bridge. Taking this to be a sign from God, he entered a monastery, where he pursued theology and philosophy, writing his famous Pensées. He also continued his mathematical research. He valued faith and intuition more than reason as the source of truth, declaring that “the heart has its own reasons, which reason cannot know.” SE CTI O N 9.5 \| Mathematical Induction 637 So we start with the left-hand side of the inequality and use the induction hypothesis to show that it is less than the right-hand side. For k 5 we have k 1 4 1 2 4k 4 2k 4 2k 4k 2k 2k 2 # 2k 2k1 Induction hypothesis Because 4 a 4k
Induction hypothesis Property of exponents Thus, P 2 Having proved Steps 1 and 2, we conclude by the Principle of Mathematical Induction 1 2 1 follows from P k, and this completes the induction step. k 1 1 n P is true for all natural numbers n 5. that ✎ Practice what you’ve learned: Do Exercise 21. 2 ▲ 2 9. ▼ CONCE PTS 1. Mathematical induction is a method of proving that a statement P n 2 1 that is true for all numbers n. In Step 1 we prove is true. 2. Which of the following is true about Step 2 in a proof by math- 10. 11. 12. 13. 13 33 53... 23 43 63... 1 1 2n 2 2n 1 3 n 2 3 2n 2 2n # 22 3 # 23 4 # 24... n # 2n ematical induction? (i) We prove “ 1 (ii) We prove “If k 1 P k P 1 2 is true.” 2 is true, then k 1 P 1 2 is true.” ▼ SKI LLS 3–14 ■ Use mathematical induction to prove that the formula is true for all natural numbers n. ✎ 3. 2 4 6... 2n n 1 n 1 4. 1 4 7... 3n 2 1 2 n ✎ 5. 5 8 11... 3n 2 2 1 1 2 2 3n 1 2 3n 7 2 2n 1 1 2 n 2 12 22 32.... 7. 8 2n 7 2 2 1 6 9. 13 23 33... 2n 1 n 1 2 14. 1 2 22... 2 n1 2 n 1 15. Show that n 2 n is divisible by 2 for all natural numbers n. 16. Show that 5n 1 is divisible by 4 for all natural numbers n. 17. Show that n 2 n 41 is odd for all natural numbers n. 18. Show that n 3 n 3 is divisible by 3 for all natural 4 numbers n. 19. Show that 8n 3n is divisible by 5 for all natural numbers n. 20. Show that 32n 1 is divisible by 8 for all natural numbers n. 21. Prove that n 2n for all natural numbers n. ✎ 2 2n 22. Prove that 1 23. Prove that if x 1, then
n 1 2 2 numbers n. 1 for all natural numbers n 3. 1 x n 1 nx for all natural 2 24. Show that 100n n 2 for all n 100. 3an and a1 5. Show that an 25. Let an1 ural numbers n. 5 3n1 for all nat- 26. A sequence is deﬁned recursively by an1 3an 8 and 4. Find an explicit formula for an, and then use mathemati- a1 cal induction to prove that the formula you found is true. 27. Show that x y is a factor of x n y n for all natural numbers n. k1 y Hint: x x y 28. Show that x y is a factor of x 2n1 y 2n1 for all natural k y x k1 x y. 2 2 1 1 3 4 k k numbers n. 638 CHAPTER 9 \| Sequences and Series 29–33 ■ Fn denotes the nth term of the Fibonacci sequence discussed in Section 9.1. Use mathematical induction to prove the statement. 2 n 3 n 1 for all n 2. (d) (e) n 3 n is divisible by 3 for all n 2. (f) n 3 6n 2 11n is divisible by 6 for all n 1. 2 1 29. F3n is even for all natural numbers n. 38. All Cats Are Black? What is wrong with the following 30. F1 31. F 2 1 F2 F 2 2 F3 32. F1 33. For all n 2,... Fn... F 2 n F3 F 2 3... F2n1 F2n Fn2 1 FnFn1 1 1 c n 1 0 d Fn1 Fn Fn Fn1 d c 34. Let an be the nth term of the sequence deﬁned recursively by 1 an1 1 an 1. Find a formula for an in terms of the Fibonacci and a1 numbers Fn. Prove that the formula you found is valid for all natural numbers n. 35. Let Fn be the nth term of the Fibonacci sequence. Find and prove an inequality relating n and Fn for natural numbers n. 36. Find and prove an inequality relating 100n and n 3. ▼ DISCOVE RY • DISCUSSION • WRITI NG 37. True or False? Determine whether each statement is
true or false. If you think the statement is true, prove it. If you think it is false, give an example in which it fails. 2 n 11 n (a) (b) n 2 n for all n 2. (c) 22n1 1 is divisible by 3 for all n 1. is prime for all n. p n 1 2 2 1 n denote the statement “In any group of n cats, if one cat “proof” by mathematical induction that all cats are black? Let P is black, then they are all black.” Step 1 The statement is clearly true for n 1. Step 2 Suppose that k P is true. We show that is true. k 1 P 1 2 1 2 Suppose we have a group of k 1 cats, one of whom is black; call this cat “Tadpole.” Remove some other cat (call it “Sparky”) from the group. We are left with k cats, one of whom (Tadpole) is black, so by the induction hypothesis, all k of these are black. Now put Sparky back in the group and take out Tadpole. We again have a group of k cats, all of whom—except possibly Sparky—are black. Then by the induction hypothesis, Sparky must be black too. So all k 1 cats in the original group are black. is true for all n. Since everyone has Thus, by induction seen at least one black cat, it follows that all cats are black. P n 1 2 Tadpole Sparky 9.6 The Binomial Theorem LEARNING OBJECTIVES After completing this section, you will be able to: ■ Expand powers of binomials using Pascal’s triangle ■ Find binomial coefﬁcients ■ Expand powers of binomials using the Binomial Theorem ■ Find a particular term in a binomial expansion An expression of the form a b is called a binomial. Although in principle it’s easy to raise a b to any power, raising it to a very high power would be tedious. In this section for any natural number n and then we ﬁnd a formula that gives the expansion of prove it using mathematical induction. a b 1 2 n SE CTI O N 9.6 \| The Binomial Theorem 639 a b 1 2 n, we ﬁrst look at some special cases: 2 2 2ab b 3 3
a 4 4a 5 5a 2b 3ab 3b 6a 2b 4b 10a 3 2 b 2 4ab 2 10a 3b 3 b 4 2b 3 5ab 4 b 5 2 1 ■ Expanding (a b)n To ﬁnd a pattern in the expansion of The following simple patterns emerge for the expansion of a b 2 1. There are n 1 terms, the ﬁrst being an and the last being bn. 2. The exponents of a decrease by 1 from term to term, while the exponents of b in- n : 1 crease by 1. 3. The sum of the exponents of a and b in each term is n. For instance, notice how the exponents of a and b behave in the expansion of a b 5. 2 1 The exponents of a decrease: a b 1 2 5 4 5 a 5a b1 10a b2 10a b3 5a b4 b5 2 3 1 The exponents of b increase: a b 1 2 5 a5 5a4b 10a3b 10a2b 5a1b b 4 5 1 3 2 With these observations we can write the form of the expansion of for any natural number n. For example, writing a question mark for the missing coefﬁcients, we have 2 1 a b n Óa bÔ8 a8 a7b a6b2 a5b3 a4b4 a3b5 a2b6 ab7 b8??????? To complete the expansion, we need to determine these coefﬁcients. To ﬁnd a pattern, let’s a b for the ﬁrst few values of n in a trianwrite the coefﬁcients in the expansion of gular array as shown in the following array, which is called Pascal’s triangle 10 10 The row corresponding to is called the zeroth row and is included to show the symmetry of the array. The key observation about Pascal’s triangle is the following property. 2 1 KEY PROPERTY OF PASCAL’S TRIANGLE Every entry (other than a 1) is the sum of the two entries diagonally above it. Pascal’s triangle appears in this Chinese document by Chu Shikie, dated 1303. The title reads “The Old Method Chart of the Seven
Multiplying Squares.” The triangle was rediscovered by Pascal (see page 636). 640 CHAPTER 9 \| Sequences and Series From this property it is easy to ﬁnd any row of Pascal’s triangle from the row above it. For instance, we ﬁnd the sixth and seventh rows, starting with the ﬁfth row 10 10 5 1 15 20 15 6 1 1 7 21 35 35 21 7 1 To see why this property holds, let’s consider the following expansions a5 5a4b 10a3b2 10a2b3 5ab4 b5 –– 6 a6 6a5b 15a4b2 20a3b3 15a2b4 6ab5 b6 a b 6 1 by multiplying. Notice, for inWe arrive at the expansion of 2 is obtained via this multiplicastance, that the circled term in the expansion of tion from the two circled terms above it. We get this term when the two terms above it are multiplied by b and a, respectively. Thus, its coefﬁcient is the sum of the coefﬁcients of these two terms. We will use this observation at the end of this section when we prove the Binomial Theorem. 1 a b by Having found these patterns, we can now easily obtain the expansion of any binomial, at least to relatively small powers. E X AM P L E 1 \| Expanding a Binomial Using Pascal’s Triangle Find the expansion of a b 7 using Pascal’s triangle. 1 2 ▼ SO LUTI O N The ﬁrst term in the expansion is a7, and the last term is b7. Using the fact that the exponent of a decreases by 1 from term to term and that of b increases by 1 from term to term, we have a b 1 7 a7 a6b a5b 2 a4b3 a3b4 a2b5 ab6 b7?????? 2 The appropriate coefﬁcients appear in the seventh row of Pascal’s triangle. Thus, a b 1 2 7 a7 7a6b 21a5b2 35a4b3 35a3b4 21a2b5 7ab6 b7 ✎ Practice what you’ve learned: Do Exercise 5. ▲ E X AM P L E 2 \| Expanding a
Binomial Using Pascal’s Triangle 2 3x Use Pascal’s triangle to expand. 5 1 2 ▼ SO LUTI O N We ﬁnd the expansion of for b. Using Pascal’s triangle for the coefﬁcients, we get 2 10a a b Substituting a 2 and b 3x gives 4b 10a 5 5a 5 a a b 3b 2 1 1 2 5 2b 3 5ab 4 b 5 and then substitute 2 for a and 3x 1 2 3x 4 2 1 2 2 2 5 5 5 3x 2 1 32 240x 720x 2 2 1 4 243x ✎ Practice what you’ve learned: Do Exercise 13. 10 1 1 2 1080x 3x 2 3 810x 2 10 2 1 2 1 2 2 3 5 3x 3 5 2 1 2 1 3x 2 4 3x 5 2 1 2 ▲ ■ The Binomial Coefﬁcients Although Pascal’s triangle is useful in ﬁnding the binomial expansion for reasonably small a b values of n, it isn’t practical for ﬁnding for large values of n. The reason is that the method we use for ﬁnding the successive rows of Pascal’s triangle is recursive. Thus, to ﬁnd the 100th row of this triangle, we must ﬁrst ﬁnd the preceding 99 rows. 1 2 n 4! 1 # 2 # 3 # 4 24 7 5040 10 # 10 3,628,800 SE CTI O N 9.6 \| The Binomial Theorem 641 We need to examine the pattern in the coefﬁcients more carefully to develop a formula that allows us to calculate directly any coefﬁcient in the binomial expansion. Such a formula exists, and the rest of this section is devoted to ﬁnding and proving it. However, to state this formula, we need some notation. The product of the ﬁrst n natural numbers is denoted by n! and is called n factorial: n We also deﬁne 0! as follows: 0! 1 This deﬁnition of 0! makes many formulas involving factorials shorter and easier to write. THE BINOMIAL COEFFICIENT Let n and r be nonnegative integers with r n. The binomial coef�
��cient is denoted by ( ) and is deﬁned by n r n r b a n! n r! 2 r! 1 E X AM P L E 3 \| Calculating Binomial Coefﬁcients 9! 4!5! 9! 9 4 9 4 b (a) 4 126 2 1 2 (b) 100 3 b a 100! 100 3! 2 3! 1 (c) 100 97 b a 100! 100 97 97 # 97 # 98 # 99 # 100 1 # 2 # 3 # p # 97 1 # 2 # 3 2 1 1 98 # 99 # 100 161,700 # 97 # 98 # 99 # 100 # 97 1 98 # 99 # 100 1 # 2 # 3 161,700 2 1! 2 2 ✎ Practice what you’ve learned: Do Exercises 17 and 19. ▲ Although the binomial coefﬁcient ( ) is deﬁned in terms of a fraction, all the results of Example 3 are natural numbers. In fact, ( ) is always a natural number (see Exercise 54). Notice that the binomial coefﬁcients in parts (b) and (c) of Example 3 are equal. This is a special case of the following relation, which you are asked to prove in Exercise 52 To see the connection between the binomial coefﬁcients and the binomial expansion of a b 1, let’s calculate the following binomial coefﬁcients: n 2 642 CHAPTER 9 \| Sequences and Series 5 2 b a 5! 5 2! 2 2! 1 10 10 5 3 b a 10 These are precisely the entries in the ﬁfth row of Pascal’s triangle. In fact, we can write Pascal’s triangle as follows To demonstrate that this pattern holds, we need to show that any entry in this version of Pascal’s triangle is the sum of the two entries diagonally above it. In other words, we must show that each entry satisﬁes the key property of Pascal’s triangle. We now state this property in terms of the binomial coefﬁcients. KEY PROPERTY OF THE BINOMIAL COEFFICIENTS For any nonnegative integers r and k with r k Notice that the two terms on the left side of this equation are adjacent entries in the kth row of Pascal
’s triangle and the term on the right side is the entry diagonally below them, in the k 1 st row. Thus, this equation is a restatement of the key property of Pascal’s triangle 1 in terms of the binomial coefﬁcients. A proof of this formula is outlined in Exercise 53. 2 ■ The Binomial Theorem We are now ready to state the Binomial Theorem. THE BINOMIAL THEOREM n1b a n 2 b a n2 n1 ab n n b a n b SE CTI O N 9.6 \| The Binomial Theorem 643 We prove this theorem at the end of this section. First, let’s look at some of its applications. E X AM P L E 4 \| Expanding a Binomial Using the Binomial Theorem Use the Binomial Theorem to expand x y 4. 2 1 ▼ SO LUTI O N By the Binomial Theorem 3y x 4 2 b a x 2y 2 4 3 b a 3 xy 4 4 b a 4 y Verify that It follows that x y 4 x 4 4x 3y 6x 2y 2 4xy 2 ✎ Practice what you’ve learned: Do Exercise 25. 1 3 y 4 ▲ E X AM P L E 5 \| Expanding a Binomial Using the Binomial Theorem Use the Binomial Theorem to expand 1x 1 A B 8. a b ▼ SO LUTI O N We ﬁrst ﬁnd the expansion of 1 1 for b. Using the Binomial Theorem, we have and then substitute 1x for a and a8 8 1 b a a7b 8 2 b a a6b 2 8 3 b a a 5b 3 8 4 b a a 4b 4 8 5 b a a 3b 5 8 6 b a 2b6 a 8 7 b a ab7 8 8 b a b8 Verify that 28 8 3 b a 56 8 4 b a 70 8 5 b a 56 8 6 b a 28 So a b 1 2 8 a8 8a7b 28a6b2 56a5b3 70a4b4 56a3b5 28a2b6 8ab7 b8 1x 1 Performing the substitutions a x 1/2 and b 1 gives 8 8 1 70 8 28 1 4 56 1 1 2 1 1 2 1 1 x1
/2 2 x1/2 2 7 2 x1/2 x1/2 x1/2 x1/ 56 x1/2 5 1 3 2 1 1 5 28 1 2 x1/ This simpliﬁes to 1x 1 1 2 8 x 4 8x 7/2 28x 3 56x 5/2 70x 2 56x 3/2 28x 8x 1/2 1 ✎ Practice what you’ve learned: Do Exercise 27. ▲ The Binomial Theorem can be used to ﬁnd a particular term of a binomial expansion without having to ﬁnd the entire expansion. 644 CHAPTER 9 \| Sequences and Series GENERAL TERM OF THE BINOMIAL EXPANSION The term that contains a r in the expansion of a b n is 2 1 arbnr n n r b a E X AM P L E 6 \| Finding a Particular Term in a Binomial Expansion Find the term that contains x 5 in the expansion of 2x y 20. 1 2 ▼ SO LUTI O N The term that contains x 5 is given by the formula for the general term with a 2x, b y, n 20, and r 5. So this term is a 5b15 20 15 b 15 20! 15!5! 1 ✎ Practice what you’ve learned: Do Exercise 39. 20! 20 15 15! 2x 5y a 1! 2 2 32x 5y 15 496,128x 5y 15 ▲ E X AM P L E 7 \| Finding a Particular Term in a Binomial Expansion Find the coefﬁcient of x 8 in the expansion of 2 1 x x b a 10. ▼ SO LUTI O N Both x 2 and 1/x are powers of x, so the power of x in each term of the expansion is determined by both terms of the binomial. To ﬁnd the required coefﬁcient, we ﬁrst ﬁnd the general term in the expansion. By the formula we have a x 2, b 1/x, and n 10, so the general term is 10r 10 10 10 10 r b a x 2r 1 x 1 10r 2 10 10 r b a x 3r10 Thus, the term that contains x 8 is the term in which 3r 10 Sir Isaac Newton (1642–1727) is universally regarded
as one of the giants of physics and mathematics. He is well known for discovering the laws of motion and gravity and for inventing the calculus, but he also proved the Binomial Theorem and the laws of optics, and he developed methods for solving polynomial equations to any desired accuracy. He was born on Christmas Day, a few months after the death of his father. After an unhappy childhood, he entered Cambridge University, where he learned mathematics by studying the writings of Euclid and Descartes. During the plague years of 1665 and 1666, when the university was closed, Newton thought and wrote about ideas that, once published, instantly revolutionized the sciences. Imbued with a pathological fear of criticism, he published these writings only after many years of encouragement from Edmund Halley (who discovered the now-famous comet) and other colleagues. Newton’s works brought him enormous fame and prestige. Even poets were moved to praise; Alexander Pope wrote: Nature and Nature’s Laws lay hid in Night. God said, “Let Newton be” and all was Light. Newton was far more modest about his accomplishments. He said, “I seem to have been only like a boy playing on the seashore... while the great ocean of truth lay all undiscovered before me.” Newton was knighted by Queen Anne in 1705 and was buried with great honor in Westminster Abbey. SE CTI O N 9.6 \| The Binomial Theorem 645 So the required coefﬁcient is 10 10 6 b a 10 4 b a 210 ✎ Practice what you’ve learned: Do Exercise 41. ▲ ■ Proof of the Binomial Theorem We now give a proof of the Binomial Theorem using mathematical induction. ▼ P RO O F Let P n 1 2 denote the statement n1b a n 2 b a n2b n1 ab n n b a n b Step 1 We show that P is true. But a1 a which is certainly true. is just the statement b1 1a 1b a b 1 2 1 1 1 b Step 2 We assume that k1b a is true. Thus, our induction hypothesis is a k 2 b k 1 k2b k1 ab k k b a k b is true. 2 We use this to show that k1b a k 2 b a k2b k1 ab k k b a k b d Induction hypothesis a b 1 2 k1
k1b a k 2 b a k2b k1 ab k1b a k 2 b a k2b k1 ab k k b a k b d Distributive Property k 0 b a k1 a k 1 b a kb a k 2 b a k1b 2b k1 k k b a k ab k 0 b a kb a k 1 b a k1b a 2 k 2 b a k2b ab k k b a k1 b Distributive Property k 0 b a k1 kb k1b ab k k b a k1 b Group like terms a b 1 2 k1 k 1 0 a a b Using the key property of the binomial coefﬁcients, we can write each of the expressions in square brackets as a single binomial coefﬁcient. Also, writing ) (these are equal to 1 by Exer) and ( the ﬁrst and last coefﬁcients as ( cise 50) gives... k1b a kb k1 k1 k1 0 k ab k1 a b b a b a a a b k1 But this last equation is precisely tion step. P 1 k 1 2, and this completes the induc- 646 CHAPTER 9 \| Sequences and Series Having proved Steps 1 and 2, we conclude by the Principle of Mathematical Induction ▲ that the theorem is true for all natural numbers n. 9. ▼ CONCE PTS 1. An algebraic expression of the form a b, which consists of a sum of two terms, is called a. 2. We can ﬁnd the coefﬁcients in the expansion of from the nth row of triangle. So ■a4 ■a 3b ■a 2b 2 ■ab 3 ■b 4. The binomial coefﬁcients can be calculated directly by using the formula n k b a. So 4 3 b a. 32. Find the ﬁrst three terms in the expansion of 40 x 1 x b a 33. Find the middle term in the expansion of 34. Find the ﬁfth term in the expansion of 1 35. Find the 24th term in the expansion of 36. Find the 28th term in the expansion of 1 37. Find the 100th term in the expansion of 1 38. Find the second term in the expansion of 2 1 18. 2 20. 2 25. 30. x
1 ab 100. 2 4. To expand a b n, we can use the 1 this theorem, we ﬁnd 2 Theorem. Using 25 a4 ■ ■ b a a 3b ■ ■ b a a 2b 2 ■ ■ b a ab 3 ■ ■ b a b4 ✎ ✎ ▼ SKI LLS 5–16 ■ Use Pascal’s triangle to expand the expression. 39. Find the term containing x 4 in the expansion of 40. Find the term containing y 3 in the expansion of 1 A 41. Find the term containing b8 in the expansion of 2 42. Find the term that does not contain x in the expansion of 1 x 2y 2 12 y a b 2 10. 12. B 12. 8 8x 1 2x b a 43–46 ■ Factor using the Binomial Theorem. ✎ 5. 8. 11. 14. 9. 12. 15. 2x 1 4 2 5 x 1 1 12 1x 5 b 7. 10. 13. 16 1a 1b 2x 3y 2 x 2 b 2 5 a 17–24 ■ Evaluate the expression. ✎ 17. 20. 23. 24. 6 4 b a 10 5 b a 18. 21 ✎ 19. 100 98 b a 22 25–28 ■ Use the Binomial Theorem to expand the expression. ✎ 25. ✎ 27. 1 4 x 2y 2 1 1 x b a 6 26. 28. 1 1 1 x 5 2 2A B 2 4 2 29. Find the ﬁrst three terms in the expansion of 30. Find the ﬁrst four terms in the expansion of 31. Find the last two terms in the expansion of 1 20. 30. x 2y 2 1 x 1/2 1 2 1 a2/3 a1/3 25. 2 6 B 43. 44. 45. 46. 3 y 4 4 4x x x 1 3y 6x 2b 6ab 1 10 2 x 1 2 1 3 12a 8a 8 4x x 2 4xy 4 10 1 1 3 2 2 b 2 4x 6y 6x 4y 2y 3 y 4 x 1 3 2 47–52 ■ Simplify using the Binomial Theorem. 47. 1 x h 2 h 3 x 3 48. 1 x h 2 h 4 x 4 49. Show that 1 100 2. 100 1.01 1.01 2 [Hint
: Note that 1 mial Theorem to show that the sum of the ﬁrst two terms of the expansion is greater than 2.], and use the Bino- 1 0.01 100 2 1 2 50. Show that 51. Show that 52. Show that and. for 0 r n. 53. In this exercise we prove the identity a) Write the left-hand side of this equation as the sum of two fractions. (b) Show that a common denominator of the expression that you found in part (a) is r! n r 1.! 1 (c) Add the two fractions using the common denominator in 2 part (b), simplify the numerator, and note that the resulting expression is equal to the right-hand side of the equation. is an integer for all n and for 0 r n. [Suggestion: Use induction to show that the statement is true for all n, and use Exercise 53 for the induction step.] 54. Prove that n r2 1 ▼ DISCOVE RY • DISCUSSION • WRITI NG 55. Powers of Factorials Which is larger, 101! 2 [Hint: Try factoring the expressions. Do they have any common factors?] 100! or 101 2 1 1 100? 56. Sums of Binomial Coefﬁcients Add each of the ﬁrst ﬁve rows of Pascal’s triangle, as indicated. Do you see a pattern? CHAPTER 9 \| REVIEW ▼ P R O P E RTI LAS Sequences (p. 600) A sequence is a function whose domain is the set of natural numbers. Instead of writing a(n) for the value of the sequence at n, we and we refer to this value as the nth term of the generally write sequence. Sequences are often described in list form: an, a1, a2, a3, p Partial Sums of a Sequence (pp. 605–606) a1, a2, a3, p For the sequence the ﬁrst n terms of the sequence: a2 p an a3 a1 Sn the nth partial sum is the sum of Sn The nth partial sum of a sequence can also be expressed by using sigma notation: Sn n a k1 ak Arithmetic Sequences (p. 611) An arithmetic sequence is a sequence whose terms are obtained by adding the same ﬁx
erm of the sequence is an a n 1 1 d 2 CHAPTER 9 \| Review 647 10 10 5 1? On the basis of the pattern you have found, ﬁnd the sum of the nth row Prove your result by expanding Theorem. 1 1 1 n 2 using the Binomial 57. Alternating Sums of Binomial Coefﬁcients Find the sum by ﬁnding a pattern as in Exercise 56. Prove your result by expanding using the Binomial Theorem. 1 1 n 1 2 Partial Sums of an Arithmetic Sequence (p. 613) an For the arithmetic sequence a n 1 d the nth partial sum 2 is given by either of the following equiva- 1 n Sn a a k1 3 lent formulas: k 1 1 d 4 2 1. 2. Sn n 2 3 2a n 1 d 4 2 1 a an 2 b Sn n a Geometric Sequences (p. 617) A geometric sequence is a sequence whose terms are obtained by multiplying each term by the same ﬁxed constant r to get the next term. Thus, a geometric sequence has the form a, ar, ar 2, ar 3, p The number a is the ﬁrst term of the sequence, and the number r is the common ratio. The nth term of the sequence is ar n1 an Partial Sums of a Geometric Sequence (p. 619) an For the geometric sequence the nth partial sum ar n1 Sn n a k1 ar k1 (where r 1 ) is given by Sn a 1 r n 1 r 648 CHAPTER 9 \| Sequences and Series Inﬁnite Geometric Series (p. 621) An inﬁnite geometric series is a series of the form a ar ar 2 ar 3 p ar n1 p 1 has the sum r An inﬁnite series for which 0 0 S a 1 r Amount of an Annuity (p. 627) The amount ments of size R with interest rate i per time period is given by of an annuity consisting of n regular equal pay- Af Af R 1 n 1 1 i 2 i Present Value of an Annuity (p. 628) The present value of an annuity consisting of n regular equal Ap payments of size R with interest rate i per time period is given by 1 n Ap R 1 1 i i 2 Present Value of a Future Amount
(p. 628) is to be paid in one lump sum, n time periods from If an amount now, and the interest rate per time period is i, then its present value Ap is given by Af Ap 1 i Af 1 Installment Buying (p. 629) If a loan rate i per time period, then the size R of each payment is given by is to be repaid in n regular equal payments with interest Ap 2 n R iAp 1 i n 2 1 1 ▼ CO N C E P T S U M MARY Section 9.1 ■ Find the terms of a sequence ■ Find the terms of a recursive sequence ■ Find the partial sums of a sequence ■ Use sigma notation Section 9.2 ■ Find the terms of an arithmetic sequence ■ Find the partial sums of an arithmetic sequence Section 9.3 ■ Find the terms of a geometric sequence ■ Find the partial sums of a geometric sequence ■ Find the sum of an inﬁnite geometric series Section 9.4 ■ Find the amount of an annuity ■ Find the present value of an annuity ■ Find the amount of the installment payments on a loan Section 9.5 ■ Prove a statement using the Principle of Mathematical Induction Principle of Mathematical Induction (p. 634) For each natural number n, let P(n) be a statement that depends on n. Suppose that each of the following conditions is satisﬁed. 1. P(1) is true. 2. For every natural number k, if P(k) is true, then Then P(n) is true for all natural numbers n. k 1 P 1 2 is true. Sums of Powers (p. 636) 0. 2. n a k1 n a k1 1 n n 1 n 1 k 2 2n 1 2 2 1 6 1. 3. n a k1 n a k1 Binomial Coefﬁcients (pp. 641–642) n r, If n and r are positive integers with then the binomial co- efﬁcient n r b a is deﬁned by n r b a n! n r! 2 r! 1 Binomial coefﬁcients satisfy the following properties The Binomial Theorem (p. 642 n1b n 2 b a a n2b2 p n n b a bn Review Exercises 1–6, 11(a)–14(
a) 7–10 11(c)–14(c), 37–40 37–48 Review Exercises 11, 14, 15–17, 25, 26, 30 50–52, 59 Review Exercises 12, 13, 18, 19, 21–24, 27–29, 31 49, 53–54, 60, 61 55–58 Review Exercises 62 63 64 Review Exercises 65–70 Section 9.6 ■ Expand a binomial using Pascal’s triangle ■ Find binomial coefﬁcients ■ Expand a binomial using the Binomial Theorem ■ Find a particular term in a binomial expansion ▼ E X E RC I S E S CHAPTER 9 \| Review 649 Review Exercises 75–76 71–74 77–78 79–81 1–6 ■ Find the ﬁrst four terms as well as the tenth term of the sequence with the given nth term. 27. The third term of a geometric sequence is 9, and the common ratio is. Find the ﬁfth term. 3 2 1. an 2 n n 1 3. an 1 1 n 1 2 3 n 5. an 1 2n! 2 2nn! 2. an 4. an 6. an 2n. an 2n 1, 7–10 ■ A sequence is deﬁned recursively. Find the ﬁrst seven terms of the sequence. an1 an1 a1 n an1 2an2, 23an1, a1 a1 13 1, a2 9. an 3 1 1 10. an an a1 8., 11–14 ■ The nth term of a sequence is given. (a) Find the ﬁrst ﬁve terms of the sequence. (b) Graph the terms you found in part (a). (c) Find the ﬁfth partial sum of the sequence. (d) Determine whether the series is arithmetic or geometric. Find the common difference or the common ratio. 11. an 13. an 2n 5 3n 2n1 12. an 14. an 5 2n 4 n 2 15–22 ■ The ﬁrst four terms of a sequence are given. Determine whether they can be the terms of an arithmetic sequence, a geometric sequence, or neither. If the sequence is arithmetic or geometric, ﬁnd the ﬁfth term. 15.
5, 5.5, 6, 6.5,... 17. t 3, t 2, t 1, t,... 19. t 3, t 2, t, 1,... 21,... 16. 12, 2 12, 3 12, 4 12,... 18. 20. 12, 2, 2 12, 4,... 1, 2, 2, 5 2,... 3 22. a, 1, 1 a, 1 2,... a 23. Show that 3, 6i, 12, 24i,... is a geometric sequence, and ﬁnd the common ratio. (Here i 11.) 24. Find the nth term of the geometric sequence 2, 2 2i, 4i, 4 4i, 8,... (Here i 11.) 25. The sixth term of an arithmetic sequence is 17, and the fourth term is 11. Find the second term. 26. The 20th term of an arithmetic sequence is 96, and the com- mon difference is 5. Find the nth term. 28. The second term of a geometric sequence is 10, and the ﬁfth term is 1250 27. Find the nth term. 29. A teacher makes $32,000 in his ﬁrst year at Lakeside School and gets a 5% raise each year. (a) Find a formula for his salary An in his nth year at this school. (b) List his salaries for his ﬁrst 8 years at this school. 30. A colleague of the teacher in Exercise 29, hired at the same time, makes $35,000 in her ﬁrst year, and gets a $1200 raise each year. (a) What is her salary An in her nth year at this school? (b) Find her salary in her eighth year at this school, and compare it to the salary of the teacher in Exercise 29 in his eighth year. 31. A certain type of bacteria divides every 5 s. If three of these bacteria are put into a petri dish, how many bacteria are in the dish at the end of 1 min? 32. If a1, a2, a3,... and b1, b2, b3,... are arithmetic sequences, b3,... is also an arithmetic b1, a2 b2, a
3 show that a1 sequence. 33. If a1, a2, a3,... and b1, b2, b3,... are geometric sequences, show that a1b1, a2b2, a3b3,... is also a geometric sequence. 34. (a) If a1, a2, a3,... is an arithmetic sequence, is the sequence a1 2, a2 2, a3 2,... arithmetic? (b) If a1, a2, a3,... is a geometric sequence, is the sequence 5a1, 5a2, 5a3,... geometric? 35. Find the values of x for which the sequence 6, x, 12,... is (a) arithmetic (b) geometric 36. Find the values of x and y for which the sequence 2, x, y, 17,... is (a) arithmetic 37–40 ■ Find the sum. 37. 39. 6 a k3 1 6 a k1 1 k 1 k 1 2 2 2 2k1 (b) geometric 38. 40. 2i 2i 1 3m2 4 a i1 5 a m1 41–44 ■ Write the sum without using sigma notation. Do not evaluate. 41. 43. k 1 2 2 10 a k1 1 50 a k1 3k 2k1 42. 100 a j2 1 j 1 10 44. a n1 n 22 n 650 CHAPTER 9 \| Sequences and Series 45–48 ■ Write the sum using sigma notation. Do not evaluate. 45. 3 6 9 12... 99 46. 12 22 32... 1002 # # 47. 1 23 2 24 3 25 4 26... 100 2102 # # # 48 999 # 1000 63. How much money should be invested every quarter at 12% per year, compounded quarterly, in order to have $10,000 in one year? 64. What are the monthly payments on a mortgage of $60,000 at 9% interest if the loan is to be repaid in (b) 15 years? (a) 30 years? 65–67 ■ Use mathematical induction to prove that the formula is true for all natural numbers n. 49–54 ■ Determine whether the expression is a partial sum of an arithmetic or geometric sequence. Then ﬁnd the sum. 2..
. 5 1 0.9 49. 1 50. 3 3.7 4.4... 10 0.9 2 0.9 1 2 51. 52. 53. 15 2 15 3 15... 100 15 2 1 3 3... 33 1 4 3 6 a n0 3 1 4 n 2 54. 8 a k0 k/2 7 5 1 2 55–58 ■ Find the sum of the inﬁnite geometric series. 4 25 1 2 5 8 55. 125 56. 0.1 0.01 0.001 0.0001...... 57.... 1 33/2 1 1 31/2 1 3 58. a ab 2 ab 4 ab 6... 59. The ﬁrst term of an arithmetic sequence is a 7, and the common difference is d 3. How many terms of this sequence must be added to obtain 325? 60. The sum of the ﬁrst three terms of a geometric series is 52, and the common ratio is r 3. Find the ﬁrst term. 61. A person has two parents, four grandparents, eight great- grandparents, and so on. What is the total number of a person’s ancestors in 15 generations? 62. Find the amount of an annuity consisting of 16 annual pay- ments of $1000 each into an account that pays 8% interest per year, compounded annually. 3n 2 2 1 n 1 3n 1 2 2... 1 2 1 2n 1 2 2n 1 1 65. 1 4 7... 66 2n 1 1 1 a 67 68. Show that 7 n 1 is divisible by 6 for all natural numbers n. 2 3n 2 for 1 1 n b 4. Show that an 4 and a1 69. Let an1 n 1... 2 b a a # 3an all natural numbers n. 70. Prove that the Fibonacci number F4n is divisible by 3 for all natural numbers n. 71–74 ■ Evaluate the expression. 71. 73 k0 a 5 k b 72. 74. 10 2 b a 10 6 b a 8 a k0 a 8 k b a 8 8 k b 75–78 ■ Expand the expression. 75. 77 76. 78. 1 1 5 x 2 2 2x y 4 79. Find the 20th term in the expansion of 1 2 a b 22. 80
. Find the ﬁrst three terms in the expansion of 1 81. Find the term containing A 6 in the expansion of 2 b2/3 b1/3 A 3B 20. 2 10. 1 2 ■ CHAPTER 9 \| TEST 1. Find the ﬁrst six terms and the sixth partial sum of the sequence whose nth term is an 2n 2 n. 2. A sequence is deﬁned recursively by an1 3an n, a1 2. Find the ﬁrst six terms of the sequence. 3. An arithmetic sequence begins 2, 5, 8, 11, 14,.... (a) Find the common difference d for this sequence. (b) Find a formula for the nth term an of the sequence. (c) Find the 35th term of the sequence. 4. A geometric sequence begins 12, 3, 3/4, 3/16, 3/64,.... (a) Find the common ratio r for this sequence. (b) Find a formula for the nth term an of the sequence. (c) Find the tenth term of the sequence. 1 5. The ﬁrst term of a geometric sequence is 25, and the fourth term is. 5 (a) Find the common ratio r and the ﬁfth term. (b) Find the partial sum of the ﬁrst eight terms. 6. The ﬁrst term of an arithmetic sequence is 10 and the tenth term is 2. (a) Find the common difference and the 100th term of the sequence. (b) Find the partial sum of the ﬁrst ten terms. 7. Let a1, a2, a3,... be a geometric sequence with initial term a and common ratio r. Show that a 2 1, a 2 2, a 2 3,... is also a geometric sequence by ﬁnding its common ratio. 8. Write the expression without using sigma notation, and then ﬁnd the sum. (a) 5 a n1 1 1 n2 2 (b) 6 a n3 1 1 2 n2n2 (a) 9. Find the sum. 1 2 32 3 1 1 21/2 (b) 22 33 1 2 23 34 1 23/2... 29 310... 10. Use mathematical induction to prove that for
all natural numbers n, 12 22 32... n2 n 1 n 1 2n 1 2 2 1 6 11. Expand 2x y 2 5. 2 1 12. Find the term containing x 3 in the binomial expansion of 3x 2 10. 2 1 13. A puppy weighs 0.85 lb at birth, and each week he gains 24% in weight. Let an be his weight in pounds at the end of his nth week of life. (a) Find a formula for an. (b) How much does the puppy weigh when he is six weeks old? (c) Is the sequence a1, a2, a3,... arithmetic, geometric, or neither? 651 MODELING WITH RECURSIVE SEQUENCES Many real-world processes occur in stages. Population growth can be viewed in stages— each new generation represents a new stage in population growth. Compound interest is paid in stages—each interest payment creates a new account balance. Many things that change continuously are more easily measured in discrete stages. For example, we can measure the temperature of a continuously cooling object in one-hour intervals. In this Focus we learn how recursive sequences are used to model such situations. In some cases we can get an explicit formula for a sequence from the recursion relation that deﬁnes it by ﬁnding a pattern in the terms of the sequence. Recursive Sequences as Models Suppose you deposit some money in an account that pays 6% interest compounded monthly. The bank has a deﬁnite rule for paying interest: At the end of each month the bank adds to your account % (or 0.005) of the amount in your account at that time. Let’s express this rule as follows: 1 2 amount at the end of this month amount at the end of last month 0.005 amount at the end of last month Using the Distributive Property, we can write this as amount at the end of this month 1.005 amount at the end of last month To model this statement using algebra, let A0 be the amount of the original deposit, A1 the amount at the end of the ﬁrst month, A2 the amount at the end of the second month, and so on. So An is the amount at the end of the nth month. Thus, An 1.005An1 We recognize this as a recursively deﬁned sequence—it gives us
the amount at each stage in terms of the amount at the preceding stage. 0.005An−1 A0 A1 A2 An−1 To ﬁnd a formula for An, let’s ﬁnd the ﬁrst few terms of the sequence and look for a pattern. 652 We see that in general, An 1 1.005 2 A1 A2 A3 A4 1.005A0 1.005A1 1.005A2 1.005A3 nA0. 1.005 1.005 1.005 1 1 1 2A0 3A0 4A0 2 2 2 Modeling with Recursive Sequences 653 E X AM P L E 1 \| Population Growth A certain animal population grows by 2% each year. The initial population is 5000. (a) Find a recursive sequence that models the population Pn at the end of the nth year. (b) Find the ﬁrst ﬁve terms of the sequence Pn. (c) Find a formula for Pn. ▼ SO LUTI O N (a) We can model the population using the following rule: population at the end of this year 1.02 population at the end of last year Algebraically, we can write this as the recursion relation (b) Since the initial population is 5000, we have Pn 1.02Pn1 P0 P1 P2 P3 P4 5000 1.02P0 1.02P1 1.02P2 1.02P3 1.02 1.02 1.02 1 1 1 2 2 2 5000 25000 35000 1.02 45000 2 (c) We see from the pattern exhibited in part (b) that Pn a geometric sequence, with common ratio r 1.02.) 1 1.02 1 2 n5000. (Note that Pn is ▲ E X AM P L E 2 \| Daily Drug Dose A patient is to take a 50-mg pill of a certain drug every morning. It is known that the body eliminates 40% of the drug every 24 hours. (a) Find a recursive sequence that models the amount An of the drug in the patient’s body after each pill is taken. (b) Find the ﬁrst four terms of the sequence An. (c) Find a formula for An. (d) How much of the drug remains in the patient’
s body after 5 days? How much will accumulate in his system after prolonged use? ▼ SO LUTI O N (a) Each morning 60% of the drug remains in his system plus he takes an additional 50 mg (his daily dose). amount of drug this morning 0.6 amount of drug yesterday morning 50 mg 654 Focus on Modeling We can express this as a recursion relation An 0.6An1 50 50 1 50 A1 A2 A0 (b) Since the initial dose is 50 mg, we have 50 0.6A0 0.6A1 0.62 50 1 0.6A2 0.63 50 50 0.6 50 0.6 0.6 50 50 2 1 0.62 0.6 1 50 0.6 0.62 50 2 1 0.6 50 1 1 0.63 0.62 0.6 1 2 0.6 50 1 50 2 0.62 3 50 A3 1 2 2 2 3 1 50 4 2 50 50 4 2 50 0.6 50 1 50 50 2 1 2 (c) From the pattern in part (b) we see that Plot1 Plot2 Plot3 Min=0 u( )=125(1-.6^( +1)) An 50 1 1 0.6 0.62... 0.6n 1 0.6n1 1 0.6 b a 50 Enter sequence 125 1 0.6n1 1 2 2 Sum of a geometric sequence: n1 1 r 1 r b a Sn a Simplify 150 0 Graph sequence 16 FIGURE 1 A5 (d) To ﬁnd the amount remaining after 5 days, we substitute n 5 and get. 125 To ﬁnd the amount remaining after prolonged use, we let n become large. As n 0 as n q (see Section 5.1). So as 1 0.651 119 mg 1 2 gets large, 0.6n approaches 0. That is, 0.6n n q, An 125 1 0.6n1 125 1 0 125 2 Thus, after prolonged use the amount of drug in the patient’s system approaches 125 mg (see Figure 1, where we have used a graphing calculator to graph the sequence). 1 1 2 ▲ Problems 1. Retirement Accounts Many college professors keep retirement savings with TIAA, the largest annuity program in the world. Interest on these accounts is compounded and credited daily. Professor Brown has $275,000 on
deposit with TIAA at the start of 2006 and receives 3.65% interest per year on his account. (a) Find a recursive sequence that models the amount An in his account at the end of the nth day of 2006. (b) Find the ﬁrst eight terms of the sequence An, rounded to the nearest cent. (c) Find a formula for An. 2. Fitness Program Sheila decides to embark on a swimming program as the best way to maintain cardiovascular health. She begins by swimming 5 min on the ﬁrst day, then adds 1 (a) Find a recursive formula for the number of minutes Tn that she swims on the nth day of min every day after that. 1 2 her program. (b) Find the ﬁrst 6 terms of the sequence Tn. (c) Find a formula for Tn. What kind of sequence is this? (d) On what day does Sheila attain her goal of swimming at least 65 min a day? (e) What is the total amount of time she will have swum after 30 days? Modeling with Recursive Sequences 655 3. Monthly Savings Program Alice opens a savings account paying 3% interest per year, compounded monthly. She begins by depositing $100 at the start of the ﬁrst month and adds $100 at the end of each month, when the interest is credited. (a) Find a recursive formula for the amount An in her account at the end of the nth month. (Include the interest credited for that month and her monthly deposit.) (b) Find the ﬁrst ﬁve terms of the sequence An. (c) Use the pattern you observed in (b) to ﬁnd a formula for An. most easily, it’s best not to simplify the terms too much.] (d) How much has she saved after 5 years? [Hint: To ﬁnd the pattern 4. Stocking a Fish Pond A pond is stocked with 4000 trout, and through reproduction the population increases by 20% per year. Find a recursive sequence that models the trout population Pn at the end of the nth year under each of the following circumstances. Find the trout population at the end of the ﬁfth year in each case. (a) The trout population changes only because of reproduction. (b) Each year 600 trout are harvested. (c) Each year 250 additional
trout are introduced into the pond. (d) Each year 10% of the trout are harvested, and 300 additional trout are introduced into the pond. 5. Pollution A chemical plant discharges 2400 tons of pollutants every year into an adjacent lake. Through natural runoff, 70% of the pollutants contained in the lake at the beginning of the year are expelled by the end of the year. (a) Explain why the following sequence models the amount An of the pollutant in the lake at the end of the nth year that the plant is operating. An 0.30An1 (b) Find the ﬁrst ﬁve terms of the sequence An. (c) Find a formula for An. (d) How much of the pollutant remains in the lake after 6 years? How much will remain after 2400 the plant has been operating a long time? (e) Verify your answer to part (d) by graphing An with a graphing calculator for n 1 to n 20. 6. Annual Savings Program Ursula opens a one-year CD that yields 5% interest per year. She begins with a deposit of $5000. At the end of each year when the CD matures, she reinvests at the same 5% interest rate, also adding 10% to the value of the CD from her other savings. (So for example, after the ﬁrst year her CD has earned 5% of $5000 in interest, for a value of $5250 at maturity. She then adds 10%, or $525, bringing the total value of her renewed CD to $5775.) (a) Find a recursive formula for the amount Un in Ursula’s CD when she reinvests at the end of the nth year. (b) Find the ﬁrst 5 terms of the sequence Un. Does this appear to be a geometric sequence? (c) Use the pattern you observed in (b) to ﬁnd a formula for Un. (d) How much has she saved after 10 years? 7. Annual Savings Program Victoria opens a one-year CD with a 5% annual interest yield at the same time as her friend Ursula in Problem 6. She also starts with an initial deposit of $5000. However, Victoria decides to add $500 to her CD when she reinvests at the end of the ﬁrst year, $1000 at the end of the second, $1500 at the end of the third
, and so on. (a) Explain why the recursive formula displayed below gives the amount Vn in Victoria’s CD when she reinvests at the end of the nth year. 1.05Vn1 Vn 500n (b) Using the Seq (“sequence”) mode on your graphing calculator, enter the sequences Un command to compare the and Vn as shown in the ﬁgure to the left. Then use the two sequences. For the ﬁrst few years, Victoria seems to be accumulating more savings than Ursula. Scroll down in the table to verify that Ursula eventually pulls ahead of Victoria in the savings race. In what year does this occur? TABLE Entering the sequences u( ) 5000 5750 6612.5 7604.4 8745 10057 11565 v( ) 5000 5750 7037.5 8889.4 11334 14401 18121 0 1 2 3 4 5 6 =0 Table of values of the sequences 656 Focus on Modeling 8. Newton’s Law of Cooling A tureen of soup at a temperature of 170F is placed on a table in a dining room in which the thermostat is set at 70F. The soup cools according to the following rule, a special case of Newton’s Law of Cooling: Each minute, the temperature of the soup declines by 3% of the difference between the soup temperature and the room temperature. (a) Find a recursive sequence that models the soup temperature Tn at the nth minute. (b) Enter the sequence Tn in your graphing calculator, and use the command to ﬁnd TABLE the temperature at 10-min increments from n 0 to n 60. (See Problem 7(b).) (c) Graph the sequence Tn. What temperature will the soup be after a long time? 9. Logistic Population Growth Simple exponential models for population growth do not take into account the fact that when the population increases, survival becomes harder for each individual because of greater competition for food and other resources. We can get a more accurate model by assuming that the birth rate is proportional to the size of the population, but the death rate is proportional to the square of the population. Using this idea, researchers ﬁnd that the number of raccoons Rn on a certain island is modeled by the following recursive sequence: Population at end of year Number of births Rn Rn1 0.
08Rn1 0.0004 Rn1 2 1 2, R0 100 Population at beginning of year Number of deaths Here n represents the number of years since observations began, R 0 is the initial population, 0.08 is the annual birth rate, and 0.0004 is a constant related to the death rate. (a) Use the TABLE command on a graphing calculator to ﬁnd the raccoon population for each year from n 1 to n 7. (b) Graph the sequence Rn. What happens to the raccoon population as n becomes large? CHAPTER 10 Counting and Probability 10.1 Counting Principles 10.2 Permutations and Combinations 10.3 Probability 10.4 Binomial Probability 10.5 Expected Value © What are the chances? Most of us are not too worried about being hit by lightning, because the chance of that happening is extremely small. But insurance companies make large payouts for lightning damage each year, so they need to know precisely the chance, or probability, of lightning striking a house. Is a hundred dollars a year a fair price to insure your house against lightning damage? What about ﬁfty dollars? Insurance companies do their probability homework carefully to make certain that their premiums are more than their payouts (see Section 10.5, Exercise 22). Of course, lightning striking a particular house is a random event, just like getting heads on a coin toss or picking the ace of spades from a shufﬂed deck. In studying probability, we will see how such random events can be modeled by using algebra. We will also encounter many other applications of probability, ranging from gauging the effectiveness of a new drug to ﬁnding the chances of winning the lottery. 657657 657 658 CHAPTER 10 \| Counting and Probability 10.1 Counting Principles LEARNING OBJECTIVE After completing this section, you will be able to: ■ Use the Fundamental Counting Principle Suppose that three towns—Ashbury, Brampton, and Carmichael—are located in such a way that two roads connect Ashbury to Brampton and three roads connect Brampton to Carmichael. p Brampton Ashbury q z x y Carmichael Route px py pz qx qy qz FIGURE 1 Tree diagram How many different routes can one take to travel from Ashbury to Carmichael via Brampton? The key to answering this question is to consider the problem in stages. At
the ﬁrst stage—from Ashbury to Brampton—there are two choices. For each of these choices there are three choices at the second stage—from Brampton to Carmichael. Thus, the number of different routes is 2 3 6. These routes are conveniently enumerated by a tree diagram as in Figure 1. The method that we used to solve this problem leads to the following principle. FUNDAMENTAL COUNTING PRINCIPLE Suppose that two events occur in order. If the ﬁrst can occur in m ways and the second in n ways (after the ﬁrst has occurred), then the two events can occur in order in m n ways. There is an immediate consequence of this principle for any number of events: If are events that occur in order and if E1 can occur in n1 ways, E2 in n2 ways, n2 E1, E2,..., Ek and so on, then the events can occur in order in... nk ways. n1 E X AM P L E 1 \| Using the Fundamental Counting Principle An ice-cream store offers three types of cones and 31 ﬂavors. How many different singlescoop ice-cream cones is it possible to buy at this store? ▼ SO LUTI O N There are two choices: type of cone and ﬂavor of ice cream. At the ﬁrst stage we choose a type of cone, and at the second stage we choose a ﬂavor. We can think of the different stages as boxes: Stage 1: Type of Cone Stage 2: Flavor SE CTI O N 10.1 \| Counting Principles 659 The ﬁrst box can be ﬁlled in three ways, and the second can be ﬁlled in 31 ways: 3 31 Stage 1 Stage 2 Thus, by the Fundamental Counting Principle there are 3 31 93 ways of choosing a single-scoop ice-cream cone at this store. ✎ Practice what you’ve learned: Do Exercise 3. ▲ E X AM P L E 2 \| Using the Fundamental Counting Principle In a certain state, automobile license plates display three letters followed by three digits. How many such plates are possible if repetition of the letters (a) is allowed? (b) is not allowed? ▼ SO LUTI O N (a) There are six choices, one for
each letter or digit on the license plate. As in the pre- ceding example, we sketch a box for each stage: 26 26 26 10 10 10 Letters Digits At the ﬁrst stage, we choose a letter (from 26 possible choices); at the second stage, we choose another letter (again from 26 choices); at the third stage, we choose another letter (26 choices); at the fourth stage, we choose a digit (from 10 possible choices); at the ﬁfth stage, we choose a digit (again from 10 choices); and at the sixth stage, we choose another digit (10 choices). By the Fundamental Counting Principle the number of possible license plates is 26 26 26 10 10 10 17,576,000 (b) If repetition of letters is not allowed, then we can arrange the choices as follows: 26 25 24 10 10 10 Letters Digits / Persi Diaconis (b. 1945) is currently professor of statistics and mathematics at Stanford University in California. He was born in New York City into a musical family and studied violin until the age of 14. At that time he left home to become a magician. He was a magician (apprentice and master) for ten years. Magic is still his passion, and if there were a professorship for magic, he would certainly qualify for such a post! His interest in card tricks led him to a study of probability and statistics. He is now one of the leading statisticians in the world. With his unusual background he approaches mathematics with an undeniable ﬂair. He says, “Statistics is the physics of numbers. Numbers seem to arise in the world in an orderly fashion. When we examine the world, the same regularities seem to appear again and again.” Among his many original contributions to mathematics is a probabilistic study of the perfect card shufﬂe. 660 CHAPTER 10 \| Counting and Probability At the ﬁrst stage, we have 26 letters to choose from, but once the ﬁrst letter has been chosen, there are only 25 letters to choose from at the second stage. Once the ﬁrst two letters have been chosen, 24 letters are left to choose from for the third stage. The digits are chosen as before. Thus, the number of possible license plates in this case is 26 25 24 10 10 10 15,600,000 ✎ Practice what you’ve learned: Do Exercise 23. ▲
E X AM P L E 3 \| Using Factorial Notation In how many different ways can a race with six runners be completed? Assume that there is no tie. ▼ SO LUTI O N There are six possible choices for ﬁrst place, ﬁve choices for second place (since only ﬁve runners are left after ﬁrst place has been decided), four choices for third place, and so on. So by the Fundamental Counting Principle the number of different ways in which this race can be completed is 6 5 4 3 2 1 6! 720 ✎ Practice what you’ve learned: Do Exercise 9. ▲ Factorial notation is explained on page 641. 10. ▼ CONCE PTS 1. The Fundamental Counting Principle says that if one event can occur in m ways and a second event can occur in n ways, then ways. So if the two events can occur in order in you have two choices for shoes and three choices for hats, then the number of different shoe-hat combinations you can wear is. 2. The Fundamental Counting Principle also applies to three or more events in order. So if you have 2 choices for shoes, 5 choices for pants, 4 choices for shirts, and 3 choices for hats, then the number of different shoe-pants-shirt-hat outﬁts you ✎ can wear is. ✎ ▼ APPLICATIONS 3. Ice-Cream Cones A vendor sells ice cream from a cart on the boardwalk. He offers vanilla, chocolate, strawberry, and pistachio ice cream, served in either a wafﬂe, sugar, or plain cone. How many different single-scoop ice-cream cones can you buy from this vendor? 4. Three-Letter Words How many three-letter “words” (strings of letters) can be formed by using the 26 letters of the alphabet if repetition of letters (a) is allowed? (b) is not allowed? 5. Three-Letter Words How many three-letter “words” (strings of letters) can be formed by using the letters WXYZ if repetition of letters (a) is allowed? (b) is not allowed? 6. Horse Race Eight horses are entered in a race. (a) How many different orders are possible for completing the race? (b) In how many different ways can ﬁrst, second, and third places be

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