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1lh23fm6b | maths | binomial-theorem | general-term | <p>If the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of $$\left(\sqrt[4]{2}+\frac{1}{\sqrt[4]{3}}\right)^{\mathrm{n}}$$ is $$\sqrt{6}: 1$$, then the third term from the beginning is :</p> | [{"identifier": "A", "content": "$$30 \\sqrt{2}$$"}, {"identifier": "B", "content": "$$60 \\sqrt{3}$$"}, {"identifier": "C", "content": "$$60 \\sqrt{2}$$"}, {"identifier": "D", "content": "$$30 \\sqrt{3}$$"}] | ["B"] | null | $$
\mathrm{T}_{r+1}={ }^n \mathrm{C}_r x^{n-r} a^r
$$
<br/><br/>$$
\frac{T_5}{T_5^{\prime}}=\frac{{ }^n C_4 \times\left((2)^{\frac{1}{4}}\right)^{n-4}\left(\frac{1}{3^{1 / 4}}\right)^4}{{ }^n C_4\left(\frac{1}{3^{1 / 4}}\right)^{n-4}\left(2^{1 / 4}\right)^4}=\sqrt{6}
$$
<br/><br/>$\left[\because r\right.$th term from e... | mcq | jee-main-2023-online-6th-april-morning-shift | 5,082 |
1lh2xsi9y | maths | binomial-theorem | general-term | <p>If the coefficient of $${x^7}$$ in $${\left( {a{x^2} + {1 \over {2bx}}} \right)^{11}}$$ and $${x^{ - 7}}$$ in $${\left( {ax - {1 \over {3b{x^2}}}} \right)^{11}}$$ are equal, then :</p> | [{"identifier": "A", "content": "$$243ab = 64$$"}, {"identifier": "B", "content": "$$32ab = 729$$"}, {"identifier": "C", "content": "$$64ab = 243$$"}, {"identifier": "D", "content": "$$729ab = 32$$"}] | ["D"] | null | General term of $\left(a x^2+\frac{1}{2 b x}\right)^{11}$ is
<br/><br/>$$
T_{r+1}={ }^{11} C_r\left(a x^2\right)^{11-r}\left(\frac{1}{2 b x}\right)^r={ }^{11} C_r(a)^{11-r}\left(\frac{1}{2 b}\right)^r x^{22-3 r}
$$
<br/><br/>$$
\begin{array}{rlrl}
&\text { Now, } 22-3 r =7 \\\\
&\Rightarrow 15 =3 r \\\\
&\Rightarrow... | mcq | jee-main-2023-online-6th-april-evening-shift | 5,083 |
jaoe38c1lscoeie7 | maths | binomial-theorem | general-term | <p>The coefficient of $$x^{2012}$$ in the expansion of $$(1-x)^{2008}\left(1+x+x^2\right)^{2007}$$ is equal to _________.</p> | [] | null | 0 | <p>$$\begin{aligned}
& (1-x)(1-x)^{2007}\left(1+x+x^2\right)^{2007} \\
& (1-x)\left(1-x^3\right)^{2007} \\
& (1-x)\left({ }^{2007} C_0-{ }^{2007} C_1\left(x^3\right)+\ldots \ldots .\right)
\end{aligned}$$</p>
<p>General term</p>
<p>$$\begin{aligned}
& (1-x)\left((-1)^r{ }^{2007} C_r x^{3 r}\right) \\
& (-1)^{r 2007} C_... | integer | jee-main-2024-online-27th-january-evening-shift | 5,084 |
lv7v47v9 | maths | binomial-theorem | general-term | <p>If the constant term in the expansion of $$\left(1+2 x-3 x^3\right)\left(\frac{3}{2} x^2-\frac{1}{3 x}\right)^9$$ is $$\mathrm{p}$$, then $$108 \mathrm{p}$$ is equal to ________.</p> | [] | null | 54 | <p>$$\text { General term of }\left(\frac{3}{2} x^2-\frac{1}{3 x}\right)^9$$</p>
<p>$$T_{r+1}={ }^9 C_r\left(\frac{3}{2} x^2\right)^{9-r}\left(-\frac{1}{3 x}\right)^r={ }^9 C_r(-1)^r 3^{9-2 r} 2^{r-9} x^{18-35}$$</p>
<p>Constant term in expansion of $$\left(1+2 x-3 x^3\right)$$</p>
<p>$$\begin{aligned}
& \left(\frac{3}... | integer | jee-main-2024-online-5th-april-morning-shift | 5,086 |
lv9s1zz5 | maths | binomial-theorem | general-term | <p>If the constant term in the expansion of $$\left(\frac{\sqrt[5]{3}}{x}+\frac{2 x}{\sqrt[3]{5}}\right)^{12}, x \neq 0$$, is $$\alpha \times 2^8 \times \sqrt[5]{3}$$, then $$25 \alpha$$ is equal to :</p> | [{"identifier": "A", "content": "724"}, {"identifier": "B", "content": "742"}, {"identifier": "C", "content": "693"}, {"identifier": "D", "content": "639"}] | ["C"] | null | <p>$$\begin{aligned}
& \left(\frac{\sqrt[5]{3}}{x}+\frac{2 x}{\sqrt[5]{3}}\right)^{12} \\
& T_{r+1}={ }^{12} C r\left(\frac{\sqrt[5]{3}}{x}\right)^{12-r}\left(\frac{2 x}{\sqrt[3]{5}}\right)^r
\end{aligned}$$</p>
<p>For constant term $$-12+r+r=0$$</p>
<p>$$\begin{aligned}
& \Rightarrow \quad r=6 \\
& \therefore \quad \t... | mcq | jee-main-2024-online-5th-april-evening-shift | 5,087 |
MdI6myzXplT0kKOP | maths | binomial-theorem | integral-and-fractional-part-of-a-number | If $$n$$ is a positive integer, then $${\left( {\sqrt 3 + 1} \right)^{2n}} - {\left( {\sqrt 3 - 1} \right)^{2n}}$$ is : | [{"identifier": "A", "content": "an irrational number "}, {"identifier": "B", "content": "an odd positive integer "}, {"identifier": "C", "content": "an even positive integer "}, {"identifier": "D", "content": "a rational number other than positive integers "}] | ["A"] | null | Let $${\left( {a + x} \right)^n}$$ = Odd trems(A) + Even terms(B)
<br><br>So $${\left( {a - x} \right)^n}$$ = Odd terms(A) - Even terms(B)
<br><br>$$\therefore$$ $${\left( {a + x} \right)^n} - {\left( {a - x} \right)^n}$$
<br><br>= (A + B) - (A - B)
<br><br>= 2B
<br><br>= 2[even terms]
<br><br>= 2[ T<sub>2</sub> + T<s... | mcq | aieee-2012 | 5,089 |
hSlHFjRByfCzOzj87g1j0 | maths | binomial-theorem | integral-and-fractional-part-of-a-number | If the fractional part of the number $$\left\{ {{{{2^{403}}} \over {15}}} \right\} is \, {k \over {15}}$$, then k is equal to :
| [{"identifier": "A", "content": "8"}, {"identifier": "B", "content": "14"}, {"identifier": "C", "content": "6"}, {"identifier": "D", "content": "1"}] | ["A"] | null | $${{{2^{403}}} \over {15}}$$
<br><br>$$ = {{{2^3}\, \cdot \,{2^{400}}} \over {15}}$$
<br><br>$$ = {8 \over {15}}{\left( {16} \right)^{100}}$$
<br><br>$$ = {8 \over {15}}{\left( {15 + 1} \right)^{100}}$$
<br><br>$$ = {8 \over {15}}\left( {{}^{100}{C_0} + {}^{100}{C_1}\,15 + {}^{100}{C_2}{{\left( {15} \right)}^2} + ........ | mcq | jee-main-2019-online-9th-january-morning-slot | 5,090 |
cYkeSxybbUbNNx1ExBjgy2xukfuvxjpi | maths | binomial-theorem | integral-and-fractional-part-of-a-number | If {p} denotes the fractional part of the number p, then
<br/>$$\left\{ {{{{3^{200}}} \over 8}} \right\}$$, is equal to : | [{"identifier": "A", "content": "$${5 \\over 8}$$"}, {"identifier": "B", "content": "$${7 \\over 8}$$"}, {"identifier": "C", "content": "$${1 \\over 8}$$"}, {"identifier": "D", "content": "$${3 \\over 8}$$"}] | ["C"] | null | $$\left\{ {{{{3^{200}}} \over 8}} \right\}$$
<br><br>= $$\left\{ {{{{{\left( {{3^2}} \right)}^{100}}} \over 8}} \right\}$$
<br><br>= $$\left\{ {{{{{\left( {1 + 8} \right)}^{100}}} \over 8}} \right\}$$
<br><br>= $$\left\{ {{{1 + {}^{100}{C_1}.8 + {}^{100}{C_2}{{.8}^2} + .... + {}^{100}{C_{100}}{{.8}^{100}}} \over 8}} \r... | mcq | jee-main-2020-online-6th-september-morning-slot | 5,091 |
ldqy761e | maths | binomial-theorem | integral-and-fractional-part-of-a-number | Let $x=(8 \sqrt{3}+13)^{13}$ and $y=(7 \sqrt{2}+9)^9$. If $[t]$ denotes the greatest integer $\leq t$, then : | [{"identifier": "A", "content": "$[x]$ is odd but $[y]$ is even"}, {"identifier": "B", "content": "$[x]$ and $[y]$ are both odd"}, {"identifier": "C", "content": "$[x]+[y]$ is even"}, {"identifier": "D", "content": "$[x]$ is even but $[y]$ is odd"}] | ["C"] | null | <p>If $${I_1} + f = {(8\sqrt 3 + 13)^{13}},f' = {(8\sqrt 3 - 13)^{13}}$$</p>
<p>$${I_1} + f - f'=$$ Even</p>
<p>$${I_1} = $$ Even</p>
<p>$${I_2} + f - f' = {(7\sqrt 2 + 9)^9} + {(7\sqrt 2 - 9)^9}$$</p>
<p>= Even</p>
<p>$${I_2} = $$ Even</p> | mcq | jee-main-2023-online-30th-january-evening-shift | 5,092 |
AcHYxPg1DlPdQGa51uedV | maths | binomial-theorem | middle-term | The sum of the real values of x for which the middle term in the binomial expansion of $${\left( {{{{x^3}} \over 3} + {3 \over x}} \right)^8}$$ equals 5670 is : | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "8"}, {"identifier": "C", "content": "6"}, {"identifier": "D", "content": "4"}] | ["A"] | null | $${T_5} = {}^8{C_4}{{{x^{12}}} \over {81}} \times {{81} \over {{x^4}}} = 5670$$
<br><br>$$ \Rightarrow 70{x^8} = 5670$$
<br><br>$$ \Rightarrow x = \pm \sqrt 3 $$ | mcq | jee-main-2019-online-11th-january-morning-slot | 5,094 |
1krw2nhf2 | maths | binomial-theorem | middle-term | The ratio of the coefficient of the middle term in the expansion of (1 + x)<sup>20</sup> and the sum of the coefficients of two middle terms in expansion of (1 + x)<sup>19</sup> is _____________. | [] | null | 1 | Coeff. of middle term in (1 + x)<sup>20</sup> = $${}^{20}{C_{10}}$$ & Sum of coeff. of two middle terms in (1 + x)<sup>19</sup> = $${}^{19}{C_{9}}$$ + $${}^{19}{C_{10}}$$<br><br>So required ratio = $${{{}^{20}{C_{10}}} \over {^{19}{C_9}{ + ^{19}}{C_{10}}}} = {{^{20}{C_{10}}} \over {^{20}{C_{10}}}} = 1$$ | integer | jee-main-2021-online-25th-july-morning-shift | 5,095 |
1ldsfiyxc | maths | binomial-theorem | middle-term | <p>Let K be the sum of the coefficients of the odd powers of $$x$$ in the expansion of $$(1+x)^{99}$$. Let $$a$$ be the middle term in the expansion of $${\left( {2 + {1 \over {\sqrt 2 }}} \right)^{200}}$$. If $${{{}^{200}{C_{99}}K} \over a} = {{{2^l}m} \over n}$$, where m and n are odd numbers, then the ordered pair $... | [{"identifier": "A", "content": "(50, 101)"}, {"identifier": "B", "content": "(50, 51)"}, {"identifier": "C", "content": "(51, 101)"}, {"identifier": "D", "content": "(51, 99)"}] | ["A"] | null | <p>$$K = {2^{98}}$$</p>
<p>$$a = {}^{200}{C_{100}}\,{2^{50}}$$</p>
<p>$$\therefore$$ $${{{}^{200}{C_{99}}\,.\,{2^{98}}} \over {{}^{200}{C_{100}}\,.\,{2^{50}}}} = {{{2^l}m} \over n}$$</p>
<p>$$ \Rightarrow {{100} \over {101}}\,.\,{2^{48}} = {{{2^l}m} \over n}$$</p>
<p>$$ \Rightarrow {{25} \over {101}}\,.\,{2^{50}} = {{{... | mcq | jee-main-2023-online-29th-january-evening-shift | 5,097 |
mSr47psylDpWgZwl9xjgy2xukf8zzatb | maths | binomial-theorem | multinomial-theorem | Let $${\left( {2{x^2} + 3x + 4} \right)^{10}} = \sum\limits_{r = 0}^{20} {{a_r}{x^r}} $$<br/><br/>
Then $${{{a_7}} \over {{a_{13}}}}$$ is equal to ______. | [] | null | 8 | <b>Note : </b> <b>Multinomial Theorem : </b>
<br><br>The general term of $${\left( {{x_1} + {x_2} + ... + {x_n}} \right)^n}$$ the expansion is
<br><br>$${{n!} \over {{n_1}!{n_2}!...{n_n}!}}x_1^{{n_1}}x_2^{{n_2}}...x_n^{{n_n}}$$
<br><br>where n<sub>1</sub> + n<sub>2</sub> + ..... + n<sub>n</sub> = n
<br><br>Here, in $$... | integer | jee-main-2020-online-4th-september-morning-slot | 5,098 |
1ktkde0y0 | maths | binomial-theorem | multinomial-theorem | If the coefficient of a<sup>7</sup>b<sup>8</sup> in the expansion of (a + 2b + 4ab)<sup>10</sup> is K.2<sup>16</sup>, then K is equal to _____________. | [] | null | 315 | $${{10!} \over {\alpha !\beta !\gamma !}}{a^\alpha }{(2b)^\beta }.{(4ab)^\gamma }$$<br><br>$${{10!} \over {\alpha !\beta !\gamma !}}{a^{\alpha + \gamma }}.\,{b^{\beta + \gamma }}\,.\,{2^\beta }\,.\,{4^\gamma }$$<br><br>$$\alpha + \beta + \gamma = 10$$ ..... (1)<br><br>$$\alpha + \gamma = 7$$ .... (2)<br><br>$$\b... | integer | jee-main-2021-online-31st-august-evening-shift | 5,099 |
1l545j8gt | maths | binomial-theorem | multinomial-theorem | <p>If the constant term in the expansion of
<br/><br/>$${\left( {3{x^3} - 2{x^2} + {5 \over {{x^5}}}} \right)^{10}}$$ is 2<sup>k</sup>.l, where l is an odd integer, then the value of k is equal to:</p> | [{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "7"}, {"identifier": "C", "content": "8"}, {"identifier": "D", "content": "9"}] | ["D"] | null | <b>Note : </b> <b>Multinomial Theorem : </b>
<br><br>The general term of $${\left( {{x_1} + {x_2} + ... + {x_n}} \right)^n}$$ the expansion is
<br><br>$${{n!} \over {{n_1}!{n_2}!...{n_n}!}}x_1^{{n_1}}x_2^{{n_2}}...x_n^{{n_n}}$$
<br><br>where n<sub>1</sub> + n<sub>2</sub> + ..... + n<sub>n</sub> = n
<br/><br/><p>Given,... | mcq | jee-main-2022-online-29th-june-morning-shift | 5,100 |
lgnxjgya | maths | binomial-theorem | multinomial-theorem | Let $\left(a+b x+c x^{2}\right)^{10}=\sum\limits_{i=0}^{20} p_{i} x^{i}, a, b, c \in \mathbb{N}$.<br/><br/> If $p_{1}=20$ and $p_{2}=210$, then
$2(a+b+c)$ is equal to : | [{"identifier": "A", "content": "15"}, {"identifier": "B", "content": "8"}, {"identifier": "C", "content": "6"}, {"identifier": "D", "content": "12"}] | ["D"] | null | <p>We are given that $\left(a+bx+cx^2\right)^{10} = \sum_{i=0}^{20} p_i x^i$, and we are given that $p_1 = 20$ and $p_2 = 210$.<br/><br/> We need to find the value of $2(a+b+c)$.</p>
Using the multinomial theorem, we can express the expansion of $(a + bx + cx^2)^{10}$ as follows:
<br/><br/>$$
\sum\limits_{k_1+k_2+k_3=... | mcq | jee-main-2023-online-15th-april-morning-shift | 5,101 |
1lgxwe9tk | maths | binomial-theorem | multinomial-theorem | <p>The coefficient of $$x^7$$ in $${(1 - x + 2{x^3})^{10}}$$ is ___________.</p> | [] | null | 960 | Given expression is $\left(1-x+2 x^3\right)^{10}$
<br/><br/>So, general term is $\frac{10 !}{r_{1} ! r_{2} ! r_{3} !}(1)^{r_1}(-1)^{r_2} \cdot(2)^{r_3} \cdot(x)^{r_2+r_3}$
<br/><br/>Where, $r_1+r_2+r_3=10$ and $r_2+3 r_3=7$
<br/><br/>Now, for possibility,
<br/><br/>$\begin{array}{ccc}r_1 & r_2 & r_3 \\ 3 & 7 & 0 \\ 7 &... | integer | jee-main-2023-online-10th-april-morning-shift | 5,102 |
jaoe38c1lse5mmmr | maths | binomial-theorem | multinomial-theorem | <p>Let $$a$$ be the sum of all coefficients in the expansion of $$\left(1-2 x+2 x^2\right)^{2023}\left(3-4 x^2+2 x^3\right)^{2024}$$ and $$b=\lim _\limits{x \rightarrow 0}\left(\frac{\int_0^x \frac{\log (1+t)}{t^{2024}+1} d t}{x^2}\right)$$. If the equation $$c x^2+d x+e=0$$ and $$2 b x^2+a x+4=0$$ have a common root, ... | [{"identifier": "A", "content": "$$2: 1: 4$$\n"}, {"identifier": "B", "content": "$$1: 1: 4$$\n"}, {"identifier": "C", "content": "$$1: 2: 4$$\n"}, {"identifier": "D", "content": "$$4: 1: 4$$"}] | ["B"] | null | <p>Put $$x=1$$</p>
<p>$$\therefore \mathrm{a}=1$$</p>
<p>$$\mathrm{b}=\lim _\limits{\mathrm{x} \rightarrow 0} \frac{\int_\limits0^{\mathrm{x}} \frac{\ln (1+\mathrm{t})}{1+\mathrm{t}^{2024}} \mathrm{dt}}{\mathrm{x}^2}$$</p>
<p>Using L' HOPITAL Rule</p>
<p>$$\mathrm{b}=\lim _\limits{\mathrm{x} \rightarrow 0} \frac{\ln (1... | mcq | jee-main-2024-online-31st-january-morning-shift | 5,104 |
5xKBLBPPtldLjMPl | maths | binomial-theorem | negative-and-fractional-index | If $$x$$ is positive, the first negative term in the expansion of $${\left( {1 + x} \right)^{27/5}}$$ is | [{"identifier": "A", "content": "6th term "}, {"identifier": "B", "content": "7th term "}, {"identifier": "C", "content": "5th term "}, {"identifier": "D", "content": "8th term."}] | ["D"] | null | General term of $${\left( {1 + x} \right)^{n}}$$ is ($${T_{r + 1}}$$) = $${{n\left( {n - 1} \right).....\left( {n - r + 1} \right)} \over {1.2.3....r}}{x^r}$$
<br><br>$$\therefore$$ General term of $${\left( {1 + x} \right)^{27/5}}$$ = $${{{{27} \over 5}\left( {{{27} \over 5} - 1} \right).....\left( {{{27} \over 5} - r... | mcq | aieee-2003 | 5,105 |
hI198LyRc6DYt2by | maths | binomial-theorem | negative-and-fractional-index | If the expansion in powers of $$x$$ of the function $${1 \over {\left( {1 - ax} \right)\left( {1 - bx} \right)}}$$ is $${a_0} + {a_1}x + {a_2}{x^2} + {a_3}{x^3}.....$$ then $${a_n}$$ is | [{"identifier": "A", "content": "$${{{b^n} - {a^n}} \\over {b - a}}$$ "}, {"identifier": "B", "content": "$${{{a^n} - {b^n}} \\over {b - a}}$$ "}, {"identifier": "C", "content": "$${{{a^{n + 1}} - {b^{n + 1}}} \\over {b - a}}$$ "}, {"identifier": "D", "content": "$${{{b^{n + 1}} - {a^{n + 1}}} \\over {b - a}}$$ "}] | ["D"] | null | $${1 \over {\left( {1 - ax} \right)\left( {1 - bx} \right)}}$$
<br><br>= $${\left( {1 - ax} \right)^{ - 1}}{\left( {1 - bx} \right)^{ - 1}}$$
<br><br>= $$\left[ {1 + \left( { - 1} \right)\left( { - ax} \right) + {{\left( { - 1} \right)\left( { - 2} \right)} \over {1.2}}{{\left( { - ax} \right)}^2} + ...} \right]$$ -
<... | mcq | aieee-2006 | 5,107 |
1krvyf364 | maths | binomial-theorem | negative-and-fractional-index | If b is very small as compared to the value of a, so that the cube and other higher powers of $${b \over a}$$ can be neglected in the identity $${1 \over {a - b}} + {1 \over {a - 2b}} + {1 \over {a - 3b}} + ..... + {1 \over {a - nb}} = \alpha n + \beta {n^2} + \gamma {n^3}$$, then the value of $$\gamma$$ is : | [{"identifier": "A", "content": "$${{{a^2} + b} \\over {3{a^3}}}$$"}, {"identifier": "B", "content": "$${{a + b} \\over {3{a^2}}}$$"}, {"identifier": "C", "content": "$${{{b^2}} \\over {3{a^3}}}$$"}, {"identifier": "D", "content": "$${{a + {b^2}} \\over {3{a^3}}}$$"}] | ["C"] | null | $${(a - b)^{ - 1}} + {(a - 2b)^{ - 1}} + .... + {(a - nb)^{ - 1}}$$<br><br>$$ = {1 \over a}\sum\limits_{r = 1}^n {{{\left( {1 - {{rb} \over a}} \right)}^{ - 1}}} $$<br><br>$$ = {1 \over a}\sum\limits_{r = 1}^n {\left\{ {\left( {1 + {{rb} \over a} + {{{r^2}{b^2}} \over {{a^2}}}} \right) + (terms\,to\,be\,neglected)} \ri... | mcq | jee-main-2021-online-25th-july-morning-shift | 5,108 |
Dv471cd1hiN4I75F | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | $$r$$ and $$n$$ are positive integers $$\,r > 1,\,n > 2$$ and coefficient of $$\,{\left( {r + 2} \right)^{th}}$$ term and $$3{r^{th}}$$ term in the expansion of $${\left( {1 + x} \right)^{2n}}$$ are equal, then $$n$$ equals | [{"identifier": "A", "content": "$$3r$$"}, {"identifier": "B", "content": "$$3r + 1$$ "}, {"identifier": "C", "content": "$$2r$$ "}, {"identifier": "D", "content": "$$2r + 1$$"}] | ["C"] | null | $$\,{\left( {r + 2} \right)^{th}}$$ term = $${}^{2n}{C_{r+1}}{\left( x \right)^r}$$
<br><br>And coefficient of $$\,{\left( {r + 2} \right)^{th}}$$ = $${}^{2n}{C_{r+1}}$$
<br><br>$$3{r^{th}}$$ term = $${}^{2n}{C_{3r - 1}}{\left( x \right)^{3r - 1}}$$
<br><br>And coefficient of $$3{r^{th}}$$ term = $${}^{2n}{C_{3r - 1}}$... | mcq | aieee-2002 | 5,110 |
9o9lRt7vfpS59rnX | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | If $${S_n} = \sum\limits_{r = 0}^n {{1 \over {{}^n{C_r}}}} \,\,and\,\,{t_n} = \sum\limits_{r = 0}^n {{r \over {{}^n{C_r}}},\,} $$then $${{{t_{ n}}} \over {{S_n}}}$$ is equal to | [{"identifier": "A", "content": "$${{2n - 1} \\over 2}$$ "}, {"identifier": "B", "content": "$${1 \\over 2}n - 1$$ "}, {"identifier": "C", "content": "n - 1"}, {"identifier": "D", "content": "$${1 \\over 2}n$$ "}] | ["D"] | null | $${S_n} = \sum\limits_{r = 0}^n {{1 \over {{}^n{C_r}}}}$$
<br><br>=$${1 \over {{}^n{C_0}}} + {1 \over {{}^n{C_1}}} + .... + {1 \over {{}^n{C_n}}}$$
<br><br>$${t_n} = \sum\limits_{r = 0}^n {{r \over {{}^n{C_r}}}}$$
<br><br>= $${0 \over {{}^n{C_0}}} + {1 \over {{}^n{C_1}}} + {2 \over {{}^n{C_2}}}.... + {n \over {{}^n{C_n... | mcq | aieee-2004 | 5,111 |
1cwigZ8CK7EUvRnc | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | The value of $$\,{}^{50}{C_4} + \sum\limits_{r = 1}^6 {^{56 - r}} {C_3}$$ is | [{"identifier": "A", "content": "$${}^{55}{C_4}$$ "}, {"identifier": "B", "content": "$${}^{55}{C_3}$$"}, {"identifier": "C", "content": "$${}^{56}{C_3}$$"}, {"identifier": "D", "content": "$${}^{56}{C_4}$$"}] | ["D"] | null | Given, $${}^{50}{C_4} + \sum\limits_{n = 1}^6 {{}^{56 - r}{C_3}} $$
<br><br>$$ \Rightarrow $$ $${}^{50}{C_4}$$ + $${}^{55}{C_3}$$ + $${}^{54}{C_3}$$ + $${}^{53}{C_3}$$ + $${}^{52}{C_3}$$ + $${}^{51}{C_3}$$ + $${}^{50}{C_3}$$
<br><br>Arrange those this way
<br><br>$$ \Rightarrow $$ $${}^{50}{C_4}$$ + $${}^{50}{C_3}$$ + ... | mcq | aieee-2005 | 5,112 |
zCRZuqBTgvBpoM9t | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | The sum of the series $${}^{20}{C_0} - {}^{20}{C_1} + {}^{20}{C_2} - {}^{20}{C_3} + .....\, - \,.....\, + {}^{20}{C_{10}}$$ is | [{"identifier": "A", "content": "$$0$$"}, {"identifier": "B", "content": "$${}^{20}{C_{10}}$$ "}, {"identifier": "C", "content": "$$ - {}^{20}{C_{10}}$$ "}, {"identifier": "D", "content": "$${1 \\over 2}{}^{20}{C_{10}}$$ "}] | ["D"] | null | We know
<br><br>$${}^{20}{C_0} - {}^{20}{C_1} + {}^{20}{C_2} - {}^{20}{C_3} + .... + {}^{20}{C_{10}} - {}^{20}{C_{11}}+ ...... + {}^{20}{C_{20}} = 0$$
<br><br>$$ \Rightarrow $$ $$({}^{20}{C_0} - {}^{20}{C_1} + {}^{20}{C_2} - {}^{20}{C_3} + .... - {}^{20} {C_{9}})$$ $$+{}^{20}{C_{10}}$$ $$(-{}^{20}{C_{9}}+ {}^{20}{C_{8}... | mcq | aieee-2007 | 5,114 |
lYC68b3a4PYHi6TL | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | <b>Statement - 1 :</b> $$\sum\limits_{r = 0}^n {\left( {r + 1} \right)\,{}^n{C_r} = \left( {n + 2} \right){2^{n - 1}}.} $$
<br/><b>Statement - 2 :</b> $$\sum\limits_{r = 0}^n {\left( {r + 1} \right)\,{}^n{C_r}{x^r} = {{\left( {1 + x} \right)}^n} + nx{{\left( {1 + x} \right)}^{n - 1}}.} $$ | [{"identifier": "A", "content": "Statement - 1 is false, Statement - 2 is true "}, {"identifier": "B", "content": "Statement - 1 is true, Statement - 2 is true; Statement - 2 is a correct explanation for Statement - 1 "}, {"identifier": "C", "content": "Statement - 1 is true, Statement - 2 is true; Statement - 2 is not... | ["B"] | null | <b>Check Statement - 1</b>
<br><br>$$\sum\limits_{r = 0}^n {\left( {r + 1} \right)\,{}^n{C_r}}$$
<br><br>= $$\sum\limits_{r = 0}^n {r.{}^n{C_r}} $$ + $$\sum\limits_{r = 0}^n {{}^n{C_r}} $$
<br><br>= $$\sum\limits_{r = 1}^n {r.{n \over r}{}^{n - 1}{C_{r - 1}}} $$ $$ + {2^n}$$
<br><br>= $$n\sum\limits_{r = 1}^n {{}^{n - ... | mcq | aieee-2008 | 5,115 |
nvvLZmA4D6F3MenV | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | Let $${s_1} = \sum\limits_{j = 1}^{10} {j\left( {j - 1} \right){}^{10}} {C_j}$$,
<br/><br/>$${{s_2} = \sum\limits_{j = 1}^{10} {} } j.{}^{10}{C_j}$$ and <br/><br/>$${{s_3} = \sum\limits_{j = 1}^{10} {{j^2}.{}^{10}{C_j}.} }$$
<p><b>Statement-1 :</b> $${{S_3} = 55 \times {2^9}}$$.
<br/><b>Statement-2 :</b> $${{S_1} = 90... | [{"identifier": "A", "content": "Statement - 1 is true, Statement- 2 is true; Statement - 2 is not a correct explanation for Statement - 1. "}, {"identifier": "B", "content": "Statement - 1 is true, Statement-2 is false. "}, {"identifier": "C", "content": "Statement - 1 is false, Statement-2 is true. "}, {"identifier":... | ["B"] | null | <b>Note :</b>
<br><br>$$\sum\limits_{r = 0}^n {r.{}^n{C_r}} $$ = $$ = n{.2^{n - 1}}$$
<br><br>$$\sum\limits_{r = 0}^n {{r^2}.{}^n{C_r}} = n\left( {n + 1} \right){2^{n - 2}}$$
<br><br>Given that,
<br><br>$${s_1} = \sum\limits_{j = 1}^{10} {j\left( {j - 1} \right){}^{10}} {C_j}$$
<br><br>=$$\sum\limits_{j = 1}^{10} {{j^... | mcq | aieee-2010 | 5,116 |
1oOYpeLuxiC0SOFR | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | The coefficient of $${x^7}$$ in the expansion of $${\left( {1 - x - {x^2} + {x^3}} \right)^6}$$ is | [{"identifier": "A", "content": "$$-132$$ "}, {"identifier": "B", "content": "$$-144$$ "}, {"identifier": "C", "content": "$$132$$ "}, {"identifier": "D", "content": "$$144$$"}] | ["B"] | null | Given,
<br>$${\left( {1 - x - {x^2} + {x^3}} \right)^6}$$
<br><br>= $${\left[ {\left( {1 - x} \right) - {x^2}\left( {1 - x} \right)} \right]^6}$$
<br><br>= $${\left( {1 - x} \right)^6}{\left( {1 - {x^2}} \right)^6}$$
<br><br>= $$\left( {1 + {}^6{C_1}( - x) + {}^6{C_2}{{( - x)}^2} + {}^6{C_3}{{( - x)}^3} + .......} \ri... | mcq | aieee-2011 | 5,117 |
YZsTRt5kvoMcWFZQ | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | If the coefficints of $${x^3}$$ and $${x^4}$$ in the expansion of $$\left( {1 + ax + b{x^2}} \right){\left( {1 - 2x} \right)^{18}}$$ in powers of $$x$$ are both zero, then $$\left( {a,\,b} \right)$$ is equal to: | [{"identifier": "A", "content": "$$\\left( {14,{{272} \\over 3}} \\right)$$ "}, {"identifier": "B", "content": "$$\\left( {16,{{272} \\over 3}} \\right)$$"}, {"identifier": "C", "content": "$$\\left( {16,{{251} \\over 3}} \\right)$$"}, {"identifier": "D", "content": "$$\\left( {14,{{251} \\over 3}} \\right)$$ "}] | ["B"] | null | $$\left( {1 + ax + b{x^2}} \right){\left( {1 - 2x} \right)^{18}}$$
<br><br>= $${\left( {1 - 2x} \right)^{18}} + ax{\left( {1 - 2x} \right)^{18}} + b{x^2}{\left( {1 - 2x} \right)^{18}}$$
<br><br>= $$\left( {1 + ax + b{x^2}} \right)\left[ {{}^{18}{C_0} - {}^{18}{C_1}\left( {2x} \right) + {}^{18}{C_2}{{\left( {2x} \right)... | mcq | jee-main-2014-offline | 5,118 |
IfImSZkdzKK4fCbz | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | The sum of coefficients of integral power of $$x$$ in the binomial expansion $${\left( {1 - 2\sqrt x } \right)^{50}}$$ is : | [{"identifier": "A", "content": "$${1 \\over 2}\\left( {{3^{50}} - 1} \\right)$$ "}, {"identifier": "B", "content": "$${1 \\over 2}\\left( {{2^{50}} + 1} \\right)$$ "}, {"identifier": "C", "content": "$${1 \\over 2}\\left( {{3^{50}} + 1} \\right)$$"}, {"identifier": "D", "content": "$${1 \\over 2}\\left( {{3^{50}}} \\r... | ["C"] | null | $${\left( {1 - 2\sqrt x } \right)^{50}}$$
<br><br>= $${}^{50}{C_0} + {}^{50}{C_1}.\left( { - 2\sqrt x } \right) + {}^{50}{C_2}.{\left( { - 2\sqrt x } \right)^2} + ....$$
<br><br>Now we need to find out those coefficient where degree of x is integer and you can see at odd terms power of x is integer.
<br><br>Let $${\lef... | mcq | jee-main-2015-offline | 5,119 |
L23VRvsDDUiZd6Fw | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | If the number of terms in the expansion of $${\left( {1 - {2 \over x} + {4 \over {{x^2}}}} \right)^n},\,x \ne 0,$$ is 28, then the sum of the coefficients of all the terms in this expansion, is : | [{"identifier": "A", "content": "243 "}, {"identifier": "B", "content": "729 "}, {"identifier": "C", "content": "64 "}, {"identifier": "D", "content": "2187 "}] | ["B"] | null | Total no of terms in $${\left( {1 - {2 \over x} + {4 \over {{x^2}}}} \right)^n}$$ = $${}^{n + 2}{C_2}$$ = 28
<br><br>(n+2)(n+1) = 56
<br><br>$$ \Rightarrow n = 6$$
<br><br>Sum of coefficient = (1 - 2 + 4)<sup>6</sup> = 3<sup>6</sup> = 729 | mcq | jee-main-2016-offline | 5,120 |
ifASjcJfTU5dfaRe | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | The value of $$\left( {{}^{21}{C_1} - {}^{10}{C_1}} \right) + \left( {{}^{21}{C_2} - {}^{10}{C_2}} \right) + \left( {{}^{21}{C_3} - {}^{10}{C_3}} \right)$$
<br/>$$\left( {{}^{21}{C_4} - {}^{10}{C_4}} \right)$$$$ + .... + \left( {{}^{21}{C_{10}} - {}^{10}{C_{10}}} \right)$$ is | [{"identifier": "A", "content": "$${2^{21}} - {2^{10}}$$"}, {"identifier": "B", "content": "$${2^{20}} - {2^{9}}$$"}, {"identifier": "C", "content": "$${2^{20}} - {2^{10}}$$"}, {"identifier": "D", "content": "$${2^{21}} - {2^{11}}$$"}] | ["C"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265679/exam_images/bx4nbx1k931n5ip0ploo.webp" loading="lazy" alt="JEE Main 2017 (Offline) Mathematics - Binomial Theorem Question 184 English Explanation"> | mcq | jee-main-2017-offline | 5,121 |
KLIGyyeNa2wGDQBt | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | The sum of the co-efficients of all odd degree terms in the expansion of
<br/><br/>$${\left( {x + \sqrt {{x^3} - 1} } \right)^5} + {\left( {x - \sqrt {{x^3} - 1} } \right)^5}$$, $$\left( {x > 1} \right)$$ is | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "-1"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "1"}] | ["A"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266144/exam_images/sxwjinh9ykk33t08ossz.webp" loading="lazy" alt="JEE Main 2018 (Offline) Mathematics - Binomial Theorem Question 185 English Explanation"> | mcq | jee-main-2018-offline | 5,122 |
1TSVDgXFTBH4OS30sfYBP | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | If n is the degree of the polynomial,
<br/><br/>$${\left[ {{2 \over {\sqrt {5{x^3} + 1} - \sqrt {5{x^3} - 1} }}} \right]^8} + $$ $${\left[ {{2 \over {\sqrt {5{x^3} + 1} + \sqrt {5{x^3} - 1} }}} \right]^8}$$
<br/><br/>and m is the coefficient of x<sup>n</sup> in it, then the ordered pair (n, m) is equal to : | [{"identifier": "A", "content": "(24, (10)<sup>8</sup>)"}, {"identifier": "B", "content": "(8, 5(10)<sup>4</sup>)"}, {"identifier": "C", "content": "(12, (20)<sup>4</sup>)"}, {"identifier": "D", "content": "(12, 8(10)<sup>4</sup>)"}] | ["C"] | null | Given, <br>
$${\left[ {{2 \over {\sqrt {5{x^3} + 1} - \sqrt {5{x^3} - 1} }}} \right]^8}$$ + $${\left[ {{2 \over {\sqrt {5{x^3} + 1} + \sqrt {5{x^3} - 1} }}} \right]^8}$$<br><br>
= $${\left[ {{2 \over {\sqrt {5{x^3} + 1} - \sqrt {5{x^3} - 1} }} \times {{\sqrt {5{x^3} + 1} + \sqrt {5{x^3} - 1} } \over {\sqrt {5{x^3}... | mcq | jee-main-2018-online-15th-april-morning-slot | 5,123 |
BBlE646RfwQmNFEvkpwnD | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | The coefficien of x<sup>10</sup> in the expansion of (1 + x)<sup>2</sup>(1 + x<sup>2</sup>)<sup>3</sup>(1 + x<sup>3</sup>)<sup>4</sup> is equal to : | [{"identifier": "A", "content": "52"}, {"identifier": "B", "content": "56"}, {"identifier": "C", "content": "50"}, {"identifier": "D", "content": "44"}] | ["A"] | null | $$ \because $$$$\,\,\,$$ (1 + x)<sup>2</sup> = 1 + 2x + x<sup>2</sup>,
<br><br>(1 + x<sup>2</sup>)<sup>3</sup> = 1 + 3x<sup>2</sup> + 3x<sup>4</sup> + x<sup>6</sup>
<br><br>and (1 + x<sup>3</sup>)<sup>4</sup> = 1 + 4x<sup>3</sup> + 6x<sup>6</sup> + 4x<sup>9</sup> + x<sup>12</sup>
<br><br>So, the possible combination fo... | mcq | jee-main-2018-online-15th-april-evening-slot | 5,124 |
vHqL1tXHdlVCD1G4cwS1m | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | The coefficient of x<sup>2</sup> in the expansion of the product
<br/>(2$$-$$x<sup>2</sup>) .((1 + 2x + 3x<sup>2</sup>)<sup>6</sup> + (1 $$-$$ 4x<sup>2</sup>)<sup>6</sup>) is : | [{"identifier": "A", "content": "107"}, {"identifier": "B", "content": "106"}, {"identifier": "C", "content": "108"}, {"identifier": "D", "content": "155"}] | ["B"] | null | Given,
<br><br>(2 $$-$$ x<sup>2</sup>) . (1 + 2x + 3x<sup>2</sup>) <sup>6</sup> + (1 $$-$$ 4x<sup>2</sup>)<sup>6</sup>)
<br><br>Let, a = ((1 + 2x + 3x<sup>2</sup>)<sup>6</sup> + (1 $$-$$ 4x<sup>2</sup>)<sup>6</sup>)
<br><br>$$\therefore\,\,\,\,$$ Given statement becomes,
<br><br>(2 $$-$$ x<sup>2</sup>) . (a)
<br><br>... | mcq | jee-main-2018-online-16th-april-morning-slot | 5,125 |
bD89kihDYtpZYSOBmF3rsa0w2w9jxaybbxn | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | If <sup>20</sup>C<sub>1</sub> + (2<sup>2</sup>) <sup>20</sup>C<sub>2</sub> + (3<sup>2</sup>) <sup>20</sup>C<sub>3</sub> + ..... + (20<sup>2</sup>
)
<sup>20</sup>C<sub>20</sub> = A(2<sup>$$\beta $$</sup>), then the ordered pair (A, $$\beta $$) is equal to : | [{"identifier": "A", "content": "(420, 19)"}, {"identifier": "B", "content": "(420, 18)"}, {"identifier": "C", "content": "(380, 18)"}, {"identifier": "D", "content": "(380, 19)"}] | ["B"] | null | S = $${1^2}\,{}^{20}{C_1} + {2^2}\,{}^{20}{C_2} + {3^2}\,{}^{20}{C_3} + .......... + {20^2}\,{}^{20}{C_{20}}$$<br><br>
$$ \Rightarrow $$ $$\sum\limits_{r = 1}^{20} {{r^2}\,{}^{20}{C_r}} $$ <br><br>
$$ \Rightarrow $$ $$\sum\limits_{r = 1}^{20} {r\,.\left( {r.{}^{20}{C_r}} \right)} $$<br><br>
$$ \Rightarrow $$ $$20\sum\l... | mcq | jee-main-2019-online-12th-april-evening-slot | 5,126 |
2O8ZKwAonWPua4lhpj3rsa0w2w9jwxtytkr | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | If the coefficients of x<sup>2</sup>
and x<sup>3</sup>
are both zero, in the expansion of the expression (1 + ax + bx<sup>2</sup>
) (1 – 3x)<sup>15</sup> in
powers of x, then the ordered pair (a,b) is equal to : | [{"identifier": "A", "content": "(28, 861)"}, {"identifier": "B", "content": "(28, 315)"}, {"identifier": "C", "content": "(\u201321, 714)"}, {"identifier": "D", "content": "(\u201354, 315)"}] | ["B"] | null | (1 + ax + bx<sup>2</sup>)(1 – 3x)<sup>15</sup><br><br>
Co-eff. of x<sup>2</sup> = 1.<sup>15</sup>C<sub>2</sub>(–3)<sup>2</sup> + a.<sup>15</sup>C<sub>1</sub>(–3) + b.<sup>15</sup>C<sub>0</sub><br><br>
$$ = {{15 \times 14} \over 2} \times 9 - 15 \times 3a + b = 0$$ (given)<br><br>
$$ \Rightarrow $$ 945 – 45a + b = 0 ...... | mcq | jee-main-2019-online-10th-april-morning-slot | 5,127 |
ULgVxIF8IPjhg5YNaS18hoxe66ijvwukqgc | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | If some three consecutive in the binomial
expansion of (x + 1)<sup>n</sup> is powers of x are in the ratio
2 : 15 : 70, then the average of these three
coefficient is :- | [{"identifier": "A", "content": "625"}, {"identifier": "B", "content": "227"}, {"identifier": "C", "content": "964"}, {"identifier": "D", "content": "232"}] | ["D"] | null | Given $${}^n{C_{r - 1}}:{}^n{C_r}:{}^n{C_{r + 1}} = 2:15:70$$<br><br>
$${{{}^n{C_{r - 1}}} \over {{}^n{C_r}}} = {2 \over {15}}$$<br><br>
$$ \Rightarrow {r \over {n - r + 1}} = {2 \over {15}}$$<br><br>
$$ \Rightarrow 15r = 2n - 2r + 2$$<br><br>
$$ \Rightarrow 17r = 2n + 2$$ .... (i)<br><br>
Now $${{{}^n{C_r}} \over {{}... | mcq | jee-main-2019-online-9th-april-evening-slot | 5,128 |
OIyzlxPBzW0N8G5uCPqZ5 | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | The sum of the co-efficients of all even
degree terms in x in the expansion of<br/>
$${\left( {x + \sqrt {{x^3} - 1} } \right)^6}$$ + $${\left( {x - \sqrt {{x^3} - 1} } \right)^6}$$, (x > 1) is equal to: | [{"identifier": "A", "content": "32"}, {"identifier": "B", "content": "26"}, {"identifier": "C", "content": "29"}, {"identifier": "D", "content": "24"}] | ["D"] | null | Let $${\left( {a + x} \right)^n}$$ = Odd trems(A) + Even terms(B)
<br><br>So $${\left( {a - x} \right)^n}$$ = Odd terms(A) - Even terms(B)
<br><br>$$\therefore$$ $${\left( {a + x} \right)^n} - {\left( {a - x} \right)^n}$$
<br><br>= (A + B) + (A - B)
<br><br>= 2A
<br><br>= 2[odd terms]
<br><br>= 2[ T<sub>1</sub> + T<su... | mcq | jee-main-2019-online-8th-april-morning-slot | 5,129 |
sd2g3rXgKxRxJIAeLbYtU | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | Let S<sub>n</sub> = 1 + q + q<sup>2</sup> + . . . . . + q<sup>n</sup> and T<sub>n</sub> = 1 + $$\left( {{{q + 1} \over 2}} \right) + {\left( {{{q + 1} \over 2}} \right)^2}$$ + . . . . . .+ $${\left( {{{q + 1} \over 2}} \right)^n}$$ where q is a real number and q $$ \ne $$ 1. If <sup>101</sup>C<sub>1</sub> + <su... | [{"identifier": "A", "content": "202"}, {"identifier": "B", "content": "200"}, {"identifier": "C", "content": "2<sup>100</sup>"}, {"identifier": "D", "content": "2<sup>99</sup>"}] | ["C"] | null | <sup>101</sup>C<sub>1</sub> + <sup>101</sup>C<sub>2</sub>S<sub>1</sub> + . . . . . . . + <sup>101</sup>C<sub>101</sub>S<sub>100</sub>
<br><br>$$=$$ $$\alpha $$T<sub>100</sub>
<br><br><sup>101</sup>C<sub>1</sub> + <sup>101</sup>C<sub>2</sub>(1 + q) + <sup>101</sup>C<sub>3</sub>(1 + q + q<sup>2</sup>) +
<br><br> &... | mcq | jee-main-2019-online-11th-january-evening-slot | 5,130 |
rgg4nMVTcWW8M9c4pNsE7 | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | Let (x + 10)<sup>50</sup> + (x $$-$$ 10)<sup>50</sup> = a<sub>0</sub> + a<sub>1</sub>x + a<sub>2</sub>x<sup>2</sup> + . . . . + a<sub>50</sub>x<sup>50</sup>, for all x $$ \in $$ R; then $${{{a_2}} \over {{a_0}}}$$ is equal to | [{"identifier": "A", "content": "12.25 "}, {"identifier": "B", "content": "12.75"}, {"identifier": "C", "content": "12.00"}, {"identifier": "D", "content": "12.50"}] | ["A"] | null | (10 + x)<sup>50</sup> + (10 $$-$$ x)<sup>50</sup>
<br><br>$$ \Rightarrow $$ a<sub>2</sub> = 2.<sup>50</sup>C<sub>2</sub> 10<sup>48</sup>, a<sub>0</sub> = 2.10<sup>50</sup>
<br><br>$${{{a_2}} \over {{a_0}}} = {{^{50}{C_2}} \over {{{10}^2}}} = 12.25$$ | mcq | jee-main-2019-online-11th-january-evening-slot | 5,131 |
MVYyLR9wVR6I6zEOqNoEu | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | The value of r for which <sup>20</sup>C<sub>r</sub> <sup>20</sup>C<sub>0</sub> + <sup>20</sup>C<sub>r$$-$$1</sub> <sup>20</sup>C<sub>1</sub> + <sup>20</sup>C<sub>r$$-$$2</sub> <sup>20</sup>C<sub>2</sub> + . . . . .+ <sup>20</sup>C<sub>0</sub> <sup>20</sup>C<sub>r</sub> is maximum, is | [{"identifier": "A", "content": "20"}, {"identifier": "B", "content": "15"}, {"identifier": "C", "content": "10"}, {"identifier": "D", "content": "11"}] | ["A"] | null | Given sum = coefficient of x<sup>r</sup> in the expansion of
<br><br>(1 + x)<sup>20</sup>(1 + x)<sup>20</sup>,
<br><br>Which is equal to <sup>40</sup>C<sub>r</sub>
<br><br>It is maximum when r = 20 | mcq | jee-main-2019-online-11th-january-morning-slot | 5,132 |
y3wzzzbvff6SL6QK9F80V | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | If $${\sum\limits_{i = 1}^{20} {\left( {{{{}^{20}{C_{i - 1}}} \over {{}^{20}{C_i} + {}^{20}{C_{i - 1}}}}} \right)} ^3} = {k \over {21}}$$ then k is equal to | [{"identifier": "A", "content": "100"}, {"identifier": "B", "content": "200"}, {"identifier": "C", "content": "50"}, {"identifier": "D", "content": "400"}] | ["A"] | null | $${\sum\limits_{i = 1}^{20} {\left( {{{^{20}{C_{i - 1}}} \over {^{20}{C_i}{ + ^{20}}{C_{i - 1}}}}} \right)} ^3} = {k \over {21}}$$
<br><br>$$ \Rightarrow \,\,\sum\limits_{i = 1}^{20} {{{\left( {{{{}^{20}{C_{i - 1}}} \over {{}^{21}{C_i}}}} \right)}^3}} = {k \over {21}}$$
<br><br>$$ \Rightarrow \,\,\sum\limits_{i = 1}^{... | mcq | jee-main-2019-online-10th-january-morning-slot | 5,133 |
CZ6nSlHjDI3ZaVDxF8B92 | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | The coefficient of t<sup>4</sup> in the expansion of $${\left( {{{1 - {t^6}} \over {1 - t}}} \right)^3}$$ is : | [{"identifier": "A", "content": "14"}, {"identifier": "B", "content": "15"}, {"identifier": "C", "content": "10"}, {"identifier": "D", "content": "12"}] | ["B"] | null | $${\left( {{{1 - {t^6}} \over {1 - t}}} \right)^3}$$
<br><br>= (1 $$-$$ t<sup>6</sup>)<sup>3</sup> (1 $$-$$ t)<sup>$$-$$3</sup>
<br><br>= (1 $$-$$ <sup>3</sup>C<sub>1</sub>t<sup>6</sup> + <sup>3</sup>C<sub>2</sub>t<sup>12</sup> $$-$$ <sup>3</sup>C<sub>3</sub>t<sup>18</sup>) $$ \times $$ (1 $$-$$ t)<sup>$$-$$3</sup>
<... | mcq | jee-main-2019-online-9th-january-evening-slot | 5,134 |
U7i5ScdA9N27m362FQs2z | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | The sum of the series
<br/><br/>2.<sup>20</sup>C<sub>0</sub>
+ 5.<sup>20</sup>C<sub>1</sub> + 8.<sup>20</sup>C<sub>2</sub> + 11.<sup>20</sup>C<sub>3</sub> + ... +62.<sup>20</sup>C<sub>20</sub> is equal to :
| [{"identifier": "A", "content": "2<sup>25</sup>"}, {"identifier": "B", "content": "2<sup>24</sup>"}, {"identifier": "C", "content": "2<sup>26</sup>"}, {"identifier": "D", "content": "2<sup>23</sup>"}] | ["A"] | null | Here general term = (3r + 2)<sup>20</sup>C<sub>r</sub>
<br><br>$$ \therefore $$ Sum of the series = $$\sum\limits_{r = 0}^{20} {\left( {3r + 2} \right)} {}^{20}{C_r}$$
<br><br>= $$3\sum\limits_{r = 0}^{20} {r.} {}^{20}{C_r} + 2\sum\limits_{r = 0}^{20} {{}^{20}{C_r}} $$
<br><br>= 3 $$ \times $$ 20$$ \times $$2<sup>20 - ... | mcq | jee-main-2019-online-8th-april-morning-slot | 5,135 |
Cq5FMsIGax6YUnb7Jrjgy2xukfajwxje | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | If for some positive integer n, the coefficients<br/> of three consecutive terms in the binomial<br/> expansion of (1 + x)<sup>n + 5</sup> are in the ratio<br/> 5 : 10 : 14, then the largest coefficient in this expansion is : | [{"identifier": "A", "content": "330"}, {"identifier": "B", "content": "792 "}, {"identifier": "C", "content": "252"}, {"identifier": "D", "content": "462"}] | ["D"] | null | Consider the three consecutive coefficients as <br><br>$$^{n + 5}{C_r},{\,^{n + 5}}{C_{r + 1}},{\,^{n + 5}}{C_{r + 2}}$$<br><br>$$ \because $$ $${{^{n + 5}{C_r}} \over {^{n + 5}{C_{r + 1}}}} = {1 \over 2}$$<br><br>$$ \Rightarrow {{r + 1} \over {n + 5 - r}} = {1 \over 2} \Rightarrow 3r = n + 3$$ ...(i)<br><br>and $${{^{... | mcq | jee-main-2020-online-4th-september-evening-slot | 5,136 |
cqGCAjsv0eMJjeXRlCjgy2xukfqgcp3c | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | The coefficient of x<sup>4</sup>
in the expansion of
<br/>(1 + x + x<sup>2</sup>
+ x<sup>3</sup>)<sup>6</sup>
in powers of x, is ______. | [] | null | 120 | (1 + x + x<sup>2</sup>
+ x<sup>3</sup>)<sup>6</sup>
<br><br>= ((1 + x) (1 + x<sup>2</sup>))<sup>6</sup>
<br><br>= (1 + x)<sup>6</sup>(1 + x<sup>2</sup>)<sup>6</sup>
<br><br>= $$\sum\limits_{r = 0}^6 {{}^6{C_r}.{x^r}} $$ $$\sum\limits_{r = 0}^6 {{}^6{C_r}.{x^{2r}}} $$
<br><br>Coefficient of x<sup>4</sup> = <sup>6</sup>... | integer | jee-main-2020-online-5th-september-evening-slot | 5,137 |
n7goxh7onsO8yp6qOojgy2xukf7fj0by | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | The value of $$\sum\limits_{r = 0}^{20} {{}^{50 - r}{C_6}} $$ is equal to: | [{"identifier": "A", "content": "$${}^{50}{C_6} - {}^{30}{C_6}$$"}, {"identifier": "B", "content": "$${}^{51}{C_7} - {}^{30}{C_7}$$"}, {"identifier": "C", "content": "$${}^{50}{C_7} - {}^{30}{C_7}$$"}, {"identifier": "D", "content": "$${}^{51}{C_7} + {}^{30}{C_7}$$"}] | ["B"] | null | $$\sum\limits_{r = 0}^{20} {} {}^{50 - r}{C_6} = {}^{50}{C_6} + {}^{49}{C_6} + {}^{48}{C_6} + .... + {}^{30}{C_6}$$<br><br>$$ = {}^{50}{C_6} + {}^{49}{C_6} + .... + {}^{31}{C_6} + ({}^{30}{C_6} + {}^{30}{C_7}) - {}^{30}{C_7}$$<br><br>$$ = {}^{50}{C_6} + {}^{49}{C_6} + .... + ({}^{31}{C_6} + {}^{31}{C_7}) - {}^{30}{C_7}... | mcq | jee-main-2020-online-4th-september-morning-slot | 5,138 |
Hs8F4tUpgblToosyEojgy2xukezfp3c8 | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | For a positive integer n,
<br/>$${\left( {1 + {1 \over x}} \right)^n}$$ is expanded
<br/>in increasing powers of x. If three consecutive
<br/>coefficients in this expansion are in the ratio,
<br/>2 : 5 : 12, then n is equal to________. | [] | null | 118 | Let, three consecutive coefficients are<br><br>
$${}^n{C_{r - 1}},{}^n{C_r},{}^n{C_{r + 1}}$$<br><br>
$${}^n{C_{r - 1}}:{}^n{C_r}:{}^n{C_{r + 1}} = 2:5:12$$<br><br>
Now, $${{{}^n{C_{r - 1}}} \over {{}^n{C_r}}} = {2 \over 5}$$<br><br>
$$ \Rightarrow 7r = 2n + 2$$ ...(i)<br><br>
$${{{}^n{C_r}} \ove... | integer | jee-main-2020-online-2nd-september-evening-slot | 5,139 |
dwSgKUsyRIPv9cmnaZ7k9k2k5e4ld1s | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | If the sum of the coefficients of all even powers of x in the product<br/> (1 + x + x<sup>2</sup>
+ ....+ x<sup>2n</sup>)(1 - x + x<sup>2</sup> - x<sup>3</sup> + ...... + x<sup>2n</sup>) is 61, then n is equal to _______. | [] | null | 30 | (1 + x + x<sup>2</sup>
+ ....+ x<sup>2n</sup>)(1 - x + x<sup>2</sup> - x<sup>3</sup> + ...... + x<sup>2n</sup>)
<br><br>= a<sub>0</sub> + a<sub>1</sub>x + a<sub>2</sub>x<sup>2</sup>
+ …..
<br><br>put x = 1
<br><br> (2n + 1)$$ \times $$1 = a<sub>0</sub> + a<sub>1</sub> + a<sub>2</sub> + …… (1)
<br><br>put x = –1
<br>... | integer | jee-main-2020-online-7th-january-morning-slot | 5,141 |
0xkP9OO8ZCBXcnBDiN1klrlf9sl | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | If $$n \ge 2$$ is a positive integer, then the sum of the series $${}^{n + 1}{C_2} + 2\left( {{}^2{C_2} + {}^3{C_2} + {}^4{C_2} + ... + {}^n{C_2}} \right)$$ is : | [{"identifier": "A", "content": "$${{n(2n + 1)(3n + 1)} \\over 6}$$"}, {"identifier": "B", "content": "$${{n(n + 1)(2n + 1)} \\over 6}$$"}, {"identifier": "C", "content": "$${{n{{(n + 1)}^2}(n + 2)} \\over {12}}$$"}, {"identifier": "D", "content": "$${{n(n - 1)(2n + 1)} \\over 6}$$"}] | ["B"] | null | $${}^{n + 1}{C_2} + 2\left( {{}^2{C_2} + {}^3{C_2} + {}^4{C_2} + ........ + {}^n{C_2}} \right)$$<br><br>$${}^{n + 1}{C_2} + 2\left( {{}^3{C_2} + {}^3{C_2} + {}^4{C_2} + ........ + {}^n{C_2}} \right)$$<br><br>use $$\left\{ {{}^n{C_{r + 1}} + {}^n{C_r} = {}^{n + 1}{C_r}} \right\}$$<br><br>$$ = {}^{n + 1}{C_2} + 2\left( {... | mcq | jee-main-2021-online-24th-february-evening-slot | 5,143 |
vof8AROKUVRxEr6t151klrmvqnc | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | For integers n and r, let $$\left( {\matrix{
n \cr
r \cr
} } \right) = \left\{ {\matrix{
{{}^n{C_r},} & {if\,n \ge r \ge 0} \cr
{0,} & {otherwise} \cr
} } \right.$$ The maximum value of k for which the sum $$\sum\limits_{i = 0}^k {\left( {\matrix{
{10} \cr
i \cr
} } \right)\l... | [] | null | 12 | As k is unbounded so maximum value is not defined.
<br><br>Question will be <b>BONUS</b>. | integer | jee-main-2021-online-24th-february-evening-slot | 5,144 |
tijFbckj71imHUIQj71kluhqs3z | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | Let m, n$$\in$$N and gcd (2, n) = 1. If $$30\left( {\matrix{
{30} \cr
0 \cr
} } \right) + 29\left( {\matrix{
{30} \cr
1 \cr
} } \right) + ...... + 2\left( {\matrix{
{30} \cr
{28} \cr
} } \right) + 1\left( {\matrix{
{30} \cr
{29} \cr
} } \right) = n{.2^m}$$, then n + m is ... | [] | null | 45 | $$30({}^{30}{C_0}) + 29({}^{30}{C_1}) + .... + 2({}^{30}{C_{28}}) + 1({}^{30}{C_{29}})$$<br><br>$$ = 30({}^{30}{C_{30}}) + 29({}^{30}{C_{29}}) + ...... + 2({}^{30}{C_2}) + 1({}^{30}{C_1})$$<br><br>$$ = \sum\limits_{r = 1}^{30} {r({}^{30}{C_r})} $$<br><br>$$ = \sum\limits_{r = 1}^{30} {r\left( {{{30} \over r}} \right)({... | integer | jee-main-2021-online-26th-february-morning-slot | 5,145 |
k0jFMPAB7RIVNOTskF1kmhxhsc1 | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | Let [ x ] denote greatest integer less than or equal to x. If for n$$\in$$N, <br/><br/>$${(1 - x + {x^3})^n} = \sum\limits_{j = 0}^{3n} {{a_j}{x^j}} $$, <br/><br/>then $$\sum\limits_{j = 0}^{\left[ {{{3n} \over 2}} \right]} {{a_{2j}} + 4} \sum\limits_{j = 0}^{\left[ {{{3n - 1} \over 2}} \right]} {{a_{2j}} + 1} $$ is eq... | [{"identifier": "A", "content": "2<sup>n $$-$$ 1</sup>"}, {"identifier": "B", "content": "n"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "1"}] | ["C"] | null | $${(1 - x + {x^3})^n} = \sum\limits_{j = 0}^{3n} {{a_j}{x^j}} $$<br><br>$$(1 - x + {x^3}) = {a_0} + {a_1}x + {a_2}{x^2} + ...... + {a_{3n}}{x^{3n}}$$<br><br>Put x = 1<br><br>$$1 = {a_0} + {a_1} + {a_2} + {a_3} + {a_4} + ........ + {a_{3n}}$$ ...... (1)<br><br>Put x = $$-$$1<br><br>$$1 = {a_0} - {a_1} + {a_2} - {a_3} + ... | mcq | jee-main-2021-online-16th-march-morning-shift | 5,146 |
VOvssU7YRUG7n4l6u61kmknei42 | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | The value of $$\sum\limits_{r = 0}^6 {\left( {{}^6{C_r}\,.\,{}^6{C_{6 - r}}} \right)} $$ is equal to : | [{"identifier": "A", "content": "924"}, {"identifier": "B", "content": "1024"}, {"identifier": "C", "content": "1124"}, {"identifier": "D", "content": "1324"}] | ["A"] | null | Given,<br><br>$$\sum\limits_{r = 0}^6 {{}^6{C_r}{}^6{C_{6 - r}}} $$
<br><br>= $${}^6{C_0}.{}^6{C_6} + {}^6{C_1}.{}^6{C_5} + ... + {}^6{C_6}.{}^6{C_0}$$
<br><br>Now,
<br><br>$$\eqalign{
& = \left( {{}^6{C_0} + {}^6{C_1}x + {}^6{C_2}{x^2} + ... + {}^6{C_6}{x^6}} \right) \cr
& \left( {{}^6{C_0} + {}^6{C_1}x... | mcq | jee-main-2021-online-17th-march-evening-shift | 5,148 |
pMbhKHUGvmcFAk55Hp1kmli04du | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | Let (1 + x + 2x<sup>2</sup>)<sup>20</sup> = a<sub>0</sub> + a<sub>1</sub>x + a<sub>2</sub>x<sup>2</sup> + .... + a<sub>40</sub>x<sup>40</sup>. Then a<sub>1</sub> + a<sub>3</sub> + a<sub>5</sub> + ..... + a<sub>37</sub> is equal to | [{"identifier": "A", "content": "2<sup>20</sup>(2<sup>20</sup> $$-$$ 21)"}, {"identifier": "B", "content": "2<sup>19</sup>(2<sup>20</sup> $$-$$ 21)"}, {"identifier": "C", "content": "2<sup>19</sup>(2<sup>20</sup> $$+$$ 21)"}, {"identifier": "D", "content": "2<sup>20</sup>(2<sup>20</sup> $$+$$ 21)"}] | ["B"] | null | $${(1 + x + 2{x^2})^{20}} = {a_0} + {a_1}x + {a_2}{x^2} + .... + {a_{40}}{x^{40}}$$<br><br>Put x = 1<br><br>$$ \Rightarrow {4^{20}} = {a_0} + {a_1} + ....... + {a_{40}}$$ ..... (i)<br><br>Put x = $$-$$1<br><br>$$ \Rightarrow {2^{20}} = {a_0} - {a_1} + ....... + - {a_{39}} + {a_{40}}$$ ..... (ii)<br><br>by (i) $$-$$ (i... | mcq | jee-main-2021-online-18th-march-morning-shift | 5,149 |
n2yplWu9PmN2Y3682V1kmm4cd5o | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | Let $${}^n{C_r}$$ denote the binomial coefficient of x<sup>r</sup> in the expansion of (1 + x)<sup>n</sup>.
If $$\sum\limits_{k = 0}^{10} {({2^2} + 3k)} {}^{10}{C_k} = \alpha {.3^{10}} + \beta {.2^{10}},\alpha ,\beta \in R$$, then $$\alpha$$ + $$\beta$$ is equal to ___________. | [] | null | 19 | $$\sum\limits_{k = 0}^{10} {({2^2} + 3k){}^{10}{C_k}} $$<br><br>$$ = 4\sum\limits_{k = 0}^{10} {{}^{10}{C_k}} + 3\sum\limits_{k = 0}^{10} {k.{}^{10}{C_k}} $$<br><br>$$ = 4({2^{10}}) + 3\sum\limits_{k = 0}^{10} {k.{{10} \over k}.{}^9{C_{k - 1}}} $$<br><br>= $$4({2^{10}}) + 3.10({2^9})$$<br><br>$$ = 4({2^{10}}) + {3.5.2... | integer | jee-main-2021-online-18th-march-evening-shift | 5,150 |
1krrogpc3 | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | For the natural numbers m, n, if $${(1 - y)^m}{(1 + y)^n} = 1 + {a_1}y + {a_2}{y^2} + .... + {a_{m + n}}{y^{m + n}}$$ and $${a_1} = {a_2} = 10$$, then the value of (m + n) is equal to : | [{"identifier": "A", "content": "88"}, {"identifier": "B", "content": "64"}, {"identifier": "C", "content": "100"}, {"identifier": "D", "content": "80"}] | ["D"] | null | $${(1 - y)^m}{(1 + y)^n} = 1 + {a_1}y + {a_2}{y^2} + .... + {a_{m + n}}{y^{m + n}}$$<br><br>Given, ($${a_1} = {a_2} = 10$$)$$(1 - my + {}^m{C_2}{y^2} + .....)(1 + ny + {}^n{C_2}{y^2} + .....) = 1 + {a_1}y + {a_2}{y^2} + ....$$<br><br>$$ \Rightarrow n - m = 10$$ ..... (i)<br><br>$$ \Rightarrow {}^m{C_2} + {}^n{C_2} - mn... | mcq | jee-main-2021-online-20th-july-evening-shift | 5,152 |
1kruc0v6x | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | The number of elements in the set {n $$\in$$ {1, 2, 3, ......., 100} | (11)<sup>n</sup> > (10)<sup>n</sup> + (9)<sup>n</sup>} is ______________. | [] | null | 96 | $${11^n} > {10^n} + {9^n}$$<br><br>$$ \Rightarrow {11^n} - {9^n} > {10^n}$$<br><br>$$ \Rightarrow {(10 + 1)^n} - {(10 - 1)^n} > {10^n}$$<br><br>$$ \Rightarrow 2\{ {}^n{C_1}{.10^{n - 1}} + {}^n{C_3}{10^{n - 10}} + {}^n{C_5}{10^{n - 5}} + .....\} > {10^n}$$
<br><br>$$ \Rightarrow $$ $${1 \over 5}\left[ {{}^n... | integer | jee-main-2021-online-22th-july-evening-shift | 5,153 |
1ktbck01t | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | If $${{}^{20}{C_r}}$$ is the co-efficient of x<sup>r</sup> in the expansion of (1 + x)<sup>20</sup>, then the value of $$\sum\limits_{r = 0}^{20} {{r^2}.{}^{20}{C_r}} $$ is equal to : | [{"identifier": "A", "content": "$$420 \\times {2^{19}}$$"}, {"identifier": "B", "content": "$$380 \\times {2^{19}}$$"}, {"identifier": "C", "content": "$$380 \\times {2^{18}}$$"}, {"identifier": "D", "content": "$$420 \\times {2^{18}}$$"}] | ["D"] | null | $$\sum\limits_{r = 0}^{20} {{r^2}.{}^{20}{C_r}} $$<br><br>$$\sum {(4(r - 1) + r).{}^{20}{C_r}} $$<br><br>$$\sum {r(r - 1).{{20 \times 19} \over {r(r - 1)}}} .{}^{18}{C_r} + r.{{20} \over r}.\sum {{}^{19}{C_{r - 1}}} $$<br><br>$$ \Rightarrow 20 \times {19.2^{18}} + {20.2^{19}}$$<br><br>$$ \Rightarrow 420 \times {2^{18}}... | mcq | jee-main-2021-online-26th-august-morning-shift | 5,155 |
1ktd4dk0v | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | Let $$\left( {\matrix{
n \cr
k \cr
} } \right)$$ denotes $${}^n{C_k}$$ and $$\left[ {\matrix{
n \cr
k \cr
} } \right] = \left\{ {\matrix{
{\left( {\matrix{
n \cr
k \cr
} } \right),} & {if\,0 \le k \le n} \cr
{0,} & {otherwise} \cr
} } \right.$$<br/><br/>If $${A_k}... | [] | null | 49 | $${A_k} = \sum\limits_{i = 0}^9 {{}^9{C_i}} {}^{12}{C_{k - i}} + \sum\limits_{i = 0}^8 {{}^8{C_i}} {}^{13}{C_{k - i}}$$<br><br>$${A_k} = {}^{21}{C_k} + {}^{21}{C_k} = 2.{}^{21}{C_k}$$<br><br>$${A_4} - {A_3} = 2\left( {{}^{21}{C_4} - {}^{21}{C_3}} \right) = 2(5985 - 1330)$$<br><br>$$190p = 2(5985 - 1330) \Rightarrow p =... | integer | jee-main-2021-online-26th-august-evening-shift | 5,156 |
1ktelqz9h | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | $$\sum\limits_{k = 0}^{20} {{{\left( {{}^{20}{C_k}} \right)}^2}} $$ is equal to : | [{"identifier": "A", "content": "$${}^{40}{C_{21}}$$"}, {"identifier": "B", "content": "$${}^{40}{C_{19}}$$"}, {"identifier": "C", "content": "$${}^{40}{C_{20}}$$"}, {"identifier": "D", "content": "$${}^{41}{C_{20}}$$"}] | ["C"] | null | $$\sum\limits_{k = 0}^{20} {{{\left( {{}^{20}{C_k}} \right)}^2}} $$
<br><br>= $${\left( {{}^{20}{C_0}} \right)^2} + {\left( {{}^{20}{C_1}} \right)^2} + {\left( {{}^{20}{C_2}} \right)^2} + .... + {\left( {{}^{20}{C_{20}}} \right)^2}$$
<br><br>= <sup>40</sup>C<sub>20</sub>
<br><br><b>Using the formula :</b>
<br><br>$${\... | mcq | jee-main-2021-online-27th-august-morning-shift | 5,157 |
1ktobm399 | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | If the sum of the coefficients in the expansion of (x + y)<sup>n</sup> is 4096, then the greatest coefficient in the expansion is _____________. | [] | null | 924 | (x + y)<sup>n</sup> $$\Rightarrow$$ 2<sup>n</sup> = 4096<br><br>2<sup>10</sup> = 1024 $$\times$$ 2<br><br>$$\Rightarrow$$ 2<sup>n</sup> = 2<sup>12</sup><br><br>2<sup>11</sup> = 2048<br><br>n = 12<br><br>2<sup>12</sup> = 4096<br><br>$${}^{12}{C_6}={{12 \times 11 \times 10 \times 9 \times 8 \times 7} \over {6 \times 5 \t... | integer | jee-main-2021-online-1st-september-evening-shift | 5,158 |
1l5662wlg | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | <p>If <br/><br/>$$\sum\limits_{k = 1}^{31} {\left( {{}^{31}{C_k}} \right)\left( {{}^{31}{C_{k - 1}}} \right) - \sum\limits_{k = 1}^{30} {\left( {{}^{30}{C_k}} \right)\left( {{}^{30}{C_{k - 1}}} \right) = {{\alpha (60!)} \over {(30!)(31!)}}} } $$, <br/><br/>where $$\alpha$$ $$\in$$ R, then the value of 16$$\alpha$$ is e... | [{"identifier": "A", "content": "1411"}, {"identifier": "B", "content": "1320"}, {"identifier": "C", "content": "1615"}, {"identifier": "D", "content": "1855"}] | ["A"] | null | <p>Given,</p>
<p>$$\sum\limits_{k = 1}^{31} {\left( {{}^{31}{C_k}} \right)\left( {{}^{31}{C_{k - 1}}} \right) - \sum\limits_{k = 1}^{30} {\left( {{}^{30}{C_k}} \right)\left( {{}^{30}{C_{k - 1}}} \right) = {{\alpha (60!)} \over {(30!)(31!)}}} } $$</p>
<p>Now,</p>
<p>$$\sum\limits_{k = 1}^{31} {\left( {{}^{31}{C_k}} \rig... | mcq | jee-main-2022-online-28th-june-morning-shift | 5,160 |
1l56ry0nx | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | <p>If the sum of the coefficients of all the positive powers of x, in the Binomial expansion of $${\left( {{x^n} + {2 \over {{x^5}}}} \right)^7}$$ is 939, then the sum of all the possible integral values of n is _________.</p> | [] | null | 57 | <p>Given, Binomial expression is</p>
<p>$$ = {\left( {{x^n} + {2 \over {{x^5}}}} \right)^7}$$</p>
<p>$$\therefore$$ General term</p>
<p>$${T_{r + 1}} = {}^7{C_r}\,.\,{({x^n})^{7 - r}}\,.\,{\left( {{2 \over {{x^5}}}} \right)^r}$$</p>
<p>$$ = {}^7{C_r}\,.\,{x^{7n - nr - 5r}}\,.\,{2^r}$$</p>
<p>For positive power of x,</p... | integer | jee-main-2022-online-27th-june-evening-shift | 5,161 |
1l58h25fw | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | If $$\left( {{}^{40}{C_0}} \right) + \left( {{}^{41}{C_1}} \right) + \left( {{}^{42}{C_2}} \right) + \,\,.....\,\, + \,\,\left( {{}^{60}{C_{20}}} \right) = {m \over n}{}^{60}{C_{20}}$$ m and n are coprime, then m + n is equal to ___________. | [] | null | 102 | <p>Here property used is</p>
<p>$${}^n{C_r} + {}^n{C_{r + 1}} = {}^{n + 1}{C_{r + 1}}$$</p>
<p>Given, $${}^{40}{C_0} + {}^{41}{C_1} + {}^{42}{C_2} + \,\,....\,\, + \,\,{}^{60}{C_{20}} = {m \over n}{}^{60}{C_{20}}$$</p>
<p>As $${}^{40}{C_0} = {}^{41}{C_0} = 1$$</p>
<p>So, we replace $${}^{40}{C_0}$$ with $${}^{41}{C_0}$... | integer | jee-main-2022-online-26th-june-evening-shift | 5,162 |
1l5ajlo7t | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | <p>Let C<sub>r</sub> denote the binomial coefficient of x<sup>r</sup> in the expansion of $${(1 + x)^{10}}$$. If for $$\alpha$$, $$\beta$$ $$\in$$ R, $${C_1} + 3.2{C_2} + 5.3{C_3} + $$ ....... upto 10 terms $$ = {{\alpha \times {2^{11}}} \over {{2^\beta } - 1}}\left( {{C_0} + {{{C_1}} \over 2} + {{{C_2}} \over 3} + \,... | [] | null | 286 | <p>Given,</p>
<p>$${C_1} + 3\,.\,2{C_2} + 5\,.\,3{C_3} + $$ ...... upto 10 terms</p>
<p>$$ = {{\alpha \,.\,{2^{11}}} \over {{2^\beta } - 1}}$$ ($${C_0} + {{{C_1}} \over 2} + {{{C_2}} \over 3}$$ + ..... upto 10 terms)</p>
<p>Now,</p>
<p>L.H.S. :-</p>
<p>$${C_1} + 3\,.\,2{C_2} + 5\,.\,3{C_3} + $$ ...... upto 10 terms</p>... | integer | jee-main-2022-online-25th-june-morning-shift | 5,163 |
1l6gjfwtl | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | <p>If the coefficients of $$x$$ and $$x^{2}$$ in the expansion of $$(1+x)^{\mathrm{p}}(1-x)^{\mathrm{q}}, \mathrm{p}, \mathrm{q} \leq 15$$, are $$-3$$ and $$-5$$ respectively, then the coefficient of $$x^{3}$$ is equal to _____________.</p> | [] | null | 23 | <p>Coefficient of x in $${(1 + x)^p}{(1 - x)^q}$$</p>
<p>$$ - {}^p{C_0}\,{}^q{C_1} + {}^p{C_1}\,{}^q{C_0} = - 3 \Rightarrow p - q = - 3$$</p>
<p>Coefficient of x<sup>2</sup> in $${(1 + x)^p}{(1 - x)^q}$$</p>
<p>$${}^p{C_0}\,{}^q{C_2} - {}^p{C_1}\,{}^q{C_1} + {}^p{C_2}\,{}^q{C_0} = - 5$$</p>
<p>$${{q(q - 1)} \over 2}... | integer | jee-main-2022-online-26th-july-morning-shift | 5,164 |
1l6hxu8ex | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | <p>$$\sum\limits_{\matrix{
{i,j = 0} \cr
{i \ne j} \cr
} }^n {{}^n{C_i}\,{}^n{C_j}} $$ is equal to</p> | [{"identifier": "A", "content": "$$2^{2 n}-{ }^{2 n} C_{n}$$"}, {"identifier": "B", "content": "$${2^{2n - 1}} - {}^{2n - 1}{C_{n - 1}}$$"}, {"identifier": "C", "content": "$$2^{2 n}-\\frac{1}{2}{ }^{2 n} C_{n}$$"}, {"identifier": "D", "content": "$${2^{2n - 1}} + {}^{2n - 1}{C_n}$$"}] | ["A"] | null | <p>$$\sum\limits_{i,\,j = 0\,\,i \ne j}^n {{}^n{C_i}\,{}^n{C_j} = \sum\limits_{i,\,j = 0}^n {{}^n{C_i}\,{}^n{C_j} - \sum\limits_{i = j}^n {{}^n{C_i}\,{}^n{C_j}} } } $$</p>
<p>$$ = \sum\limits_{j = 0}^n {{}^n{C_i}\,\sum\limits_{j = 0}^n {{}^n{C_j} - \sum\limits_{i = 0}^n {{}^n{C_i}\,{C_i}} } } $$</p>
<p>$$ = {2^n}\,.\,{... | mcq | jee-main-2022-online-26th-july-evening-shift | 5,165 |
1l6rfnnhx | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | <p>$$
\text { If } \sum\limits_{k=1}^{10} K^{2}\left(10_{C_{K}}\right)^{2}=22000 L \text {, then } L \text { is equal to }$$ ________.</p> | [] | null | 221 | <p>Given,</p>
<p>$$\sum\limits_{k = 1}^{10} {{k^2}{{\left( {{}^{10}{C_k}} \right)}^2} = 2200\,L} $$</p>
<p>$$ \Rightarrow \sum\limits_{k = 1}^{10} {{{\left( {k\,.\,{}^{10}{C_k}} \right)}^2} = 22000\,L} $$</p>
<p>$$ \Rightarrow \sum\limits_{k = 1}^{10} {{{\left( {k\,.\,{{10} \over k}\,.\,{}^9{C_{k - 1}}} \right)}^2} = 2... | integer | jee-main-2022-online-29th-july-evening-shift | 5,168 |
1ldr759il | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | <p>The coefficient of $${x^{301}}$$ in $${(1 + x)^{500}} + x{(1 + x)^{499}} + {x^2}{(1 + x)^{498}}\, + \,...\, + \,{x^{500}}$$ is :</p> | [{"identifier": "A", "content": "$${}^{500}{C_{300}}$$"}, {"identifier": "B", "content": "$${}^{501}{C_{200}}$$"}, {"identifier": "C", "content": "$${}^{500}{C_{301}}$$"}, {"identifier": "D", "content": "$${}^{501}{C_{302}}$$"}] | ["B"] | null | <p>The coefficient of $${x^{301}}$$ in
<br/><br/>$${(1 + x)^{500}} + x{(1 + x)^{499}} + {x^2}{(1 + x)^{498}}\, + \,...\, + \,{x^{500}}$$</p>
<p>$${}^{500}{C_{301}} + {}^{499}{C_{300}} + {}^{498}{C_{299}}\, + \,...\, + \,{}^{199}{C_0}$$</p>
<p>$$ = {}^{500}{C_{199}} + {}^{499}{C_{199}} + {}^{498}{C_{199}}\, + \,...\, +... | mcq | jee-main-2023-online-30th-january-morning-shift | 5,169 |
1ldv1hnql | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | <p>If $$a_r$$ is the coefficient of $$x^{10-r}$$ in the Binomial expansion of $$(1 + x)^{10}$$, then $$\sum\limits_{r = 1}^{10} {{r^3}{{\left( {{{{a_r}} \over {{a_{r - 1}}}}} \right)}^2}} $$ is equal to </p> | [{"identifier": "A", "content": "3025"}, {"identifier": "B", "content": "4895"}, {"identifier": "C", "content": "5445"}, {"identifier": "D", "content": "1210"}] | ["D"] | null | $$
\begin{aligned}
& \mathrm{a}_{\mathrm{r}}={ }^{10} \mathrm{C}_{10-\mathrm{r}}={ }^{10} \mathrm{C}_{\mathrm{r}} \\\\
& \Rightarrow \sum_{\mathrm{r}=1}^{10} \mathrm{r}^3\left(\frac{{ }^{10} \mathrm{C}_{\mathrm{r}}}{{ }^{10} \mathrm{C}_{\mathrm{r}-1}}\right)^2=\sum_{\mathrm{r}=1}^{10} \mathrm{r}^3\left(\frac{11-\mathrm... | mcq | jee-main-2023-online-25th-january-morning-shift | 5,170 |
1ldwxb38k | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | <p>If $${({}^{30}{C_1})^2} + 2{({}^{30}{C_2})^2} + 3{({}^{30}{C_3})^2}\, + \,...\, + \,30{({}^{30}{C_{30}})^2} = {{\alpha 60!} \over {{{(30!)}^2}}}$$ then $$\alpha$$ is equal to :</p> | [{"identifier": "A", "content": "30"}, {"identifier": "B", "content": "10"}, {"identifier": "C", "content": "15"}, {"identifier": "D", "content": "60"}] | ["C"] | null | $$
\begin{aligned}
& \mathrm{S}=0 \cdot\left({ }^{30} \mathrm{C}_0\right)^2+1 \cdot\left({ }^{30} \mathrm{C}_1\right)^2+2 \cdot\left({ }^{30} \mathrm{C}_2\right)^2+\ldots \ldots+30 \cdot\left({ }^{30} \mathrm{C}_{30}\right)^2 \\\\
& \mathrm{S}=30 \cdot\left({ }^{30} \mathrm{C}_0\right)^2+29 \cdot\left({ }^{30} \mathrm{... | mcq | jee-main-2023-online-24th-january-evening-shift | 5,171 |
1ldyaytnw | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | <p>The value of $$\sum\limits_{r = 0}^{22} {{}^{22}{C_r}{}^{23}{C_r}} $$ is</p> | [{"identifier": "A", "content": "$${}^{44}{C_{23}}$$"}, {"identifier": "B", "content": "$${}^{45}{C_{23}}$$"}, {"identifier": "C", "content": "$${}^{44}{C_{22}}$$"}, {"identifier": "D", "content": "$${}^{45}{C_{24}}$$"}] | ["B"] | null | <p>$$\sum\limits_{r = 0}^{22} {{}^{22}{C_r}\,.\,{}^{23}{C_r}} $$</p>
<p>$$ = \sum\limits_{r = 0}^{22} {{}^{22}{C_r}\,{}^{23}{C_{23 - r}}} $$ [using $${}^n{C_r} = {}^n{C_{n - r}}$$]</p>
<p>$$ = {}^{22}{C_0}{}^{23}{C_{23}} + {}^{22}{C_1}{}^{23}{C_{22}}\, + \,...\, + \,{}^{22}{C_{21}}{}^{23}{C_2} + {}^{22}{C_{22}}{}^{23}{... | mcq | jee-main-2023-online-24th-january-morning-shift | 5,172 |
1ldyc6w0d | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | <p>Suppose $$\sum\limits_{r = 0}^{2023} {{r^2}{}~^{2023}{C_r} = 2023 \times \alpha \times {2^{2022}}} $$. Then the value of $$\alpha$$ is ___________</p> | [] | null | 1012 | <p>Concept :</p>
<p>(1) $${}^n{C_r} = {n \over r}\,.\,{}^{n - 1}{C_{r - 1}}$$</p>
<p>Given,</p>
<p>$$\sum\limits_{r = 0}^{2023} {{r^2}\,.\,{}^{2023}{C_r}} $$</p>
<p>$$ = \sum\limits_{r = 0}^{2023} {{r^2}\,.\,{{2023} \over r}\,.{}^{2022}{C_{r - 1}}} $$</p>
<p>$$ = 2023\sum\limits_{r = 0}^{2023} {{r}\,.\,{}^{2022}{C_{r -... | integer | jee-main-2023-online-24th-january-morning-shift | 5,173 |
1lgrecwt6 | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | <p>If $$\frac{1}{n+1}{ }^{n} \mathrm{C}_{n}+\frac{1}{n}{ }^{n} \mathrm{C}_{n-1}+\ldots+\frac{1}{2}{ }^{n} \mathrm{C}_{1}+{ }^{n} \mathrm{C}_{0}=\frac{1023}{10}$$ then $$n$$ is equal to :</p> | [{"identifier": "A", "content": "9"}, {"identifier": "B", "content": "6"}, {"identifier": "C", "content": "7"}, {"identifier": "D", "content": "8"}] | ["A"] | null | $$\frac{1}{n+1}{ }^{n} \mathrm{C}_{n}+\frac{1}{n}{ }^{n} \mathrm{C}_{n-1}+\ldots+\frac{1}{2}{ }^{n} \mathrm{C}_{1}+{ }^{n} \mathrm{C}_{0}=\frac{1023}{10}$$
<br/><br/>$$
\begin{aligned}
& \Rightarrow \sum_{r=0}^n \frac{1}{r+1}{ }^n C_r=\frac{1023}{10} \\\\
& \quad\left( \because{ }^{n+1} C_{r+1}=\frac{n+1}{r+1}{ }^n C_r... | mcq | jee-main-2023-online-12th-april-morning-shift | 5,174 |
1lgrg83q5 | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | <p>The sum, of the coefficients of the first 50 terms in the binomial expansion of $$(1-x)^{100}$$, is equal to</p> | [{"identifier": "A", "content": "$${ }^{99} \\mathrm{C}_{49}$$"}, {"identifier": "B", "content": "$${ }^{101} \\mathrm{C}_{50}$$"}, {"identifier": "C", "content": "$$-{ }^{99} \\mathrm{C}_{49}$$"}, {"identifier": "D", "content": "$$-{ }^{101} \\mathrm{C}_{50}$$"}] | ["C"] | null | $$
\begin{aligned}
& \left({ }^{100} C_0-{ }^{100} C_1+{ }^{100} C_2-\ldots . .{ }^{100} C_{49}\right)+{ }^{100} C_{50} \\\\
& +\left(-{ }^{100} C_{51}+{ }^{100} C_{52}-\ldots .+{ }^{100} C_{100}\right)=0 \\\\
& \lambda_1+{ }^{100} C_{50}+\lambda_2=0 \\\\
& \lambda_1=-\frac{1}{2}{ }^{100} C_{50} \quad\left(\because \la... | mcq | jee-main-2023-online-12th-april-morning-shift | 5,175 |
1lgzzpzqm | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | <p>If the coefficients of three consecutive terms in the expansion of $$(1+x)^{n}$$ are in the ratio $$1: 5: 20$$, then the coefficient of the fourth term is</p> | [{"identifier": "A", "content": "3654"}, {"identifier": "B", "content": "1827"}, {"identifier": "C", "content": "5481"}, {"identifier": "D", "content": "2436"}] | ["A"] | null | $$
\begin{aligned}
& \text { Given: }{ }^n \mathrm{C}_{r-1}:{ }^n \mathrm{C}_r:{ }^n \mathrm{C}_{r+1} \\\\
& =1: 5: 20 \\\\
& \Rightarrow \frac{n !}{(r-1) !(n-r+1) !} \times \frac{r !(n-r) !}{n !}=\frac{1}{5} \\\\
& \Rightarrow \frac{r}{(n-r+1)}=\frac{1}{5} \\\\
& \Rightarrow 5 r=n-r+1 \\\\
& \Rightarrow n=6 r-1 ......... | mcq | jee-main-2023-online-8th-april-morning-shift | 5,176 |
1lh238a1o | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | <p>If $${ }^{2 n} C_{3}:{ }^{n} C_{3}=10: 1$$, then the ratio $$\left(n^{2}+3 n\right):\left(n^{2}-3 n+4\right)$$ is :</p> | [{"identifier": "A", "content": "$$27: 11$$"}, {"identifier": "B", "content": "$$2: 1$$"}, {"identifier": "C", "content": "$$35: 16$$"}, {"identifier": "D", "content": "$$65: 37$$"}] | ["B"] | null | $$
\begin{aligned}
& \text {We have, }{ }^{2 n} C_3:{ }^n C_3=10: 1 \\\\
& \Rightarrow \frac{{ }^{2 n} C_3}{{ }^n C_3}=\frac{10}{1} \\\\
& \Rightarrow \frac{(2 n) !}{3 !(2 n-3) !} \times \frac{3 !(n-3) !}{n !}=\frac{10}{1} \\\\
& \Rightarrow \frac{(2 n)(2 n-1)(2 n-2)}{(n)(n-1)(n-2)}=\frac{10}{1} \\\\
& \Rightarrow 4 n^... | mcq | jee-main-2023-online-6th-april-morning-shift | 5,177 |
1lh241fo2 | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | <p>The coefficient of $$x^{18}$$ in the expansion of $$\left(x^{4}-\frac{1}{x^{3}}\right)^{15}$$ is __________.</p> | [] | null | 5005 | $\begin{aligned} T_{r+1}= & { }^{15} C_r\left(x^4\right)^{15-r}\left(-\frac{1}{x^3}\right)^r={ }^{15} C_r(-1)^r x^{60-4 r-3 r} \\\\ = & { }^{15} C_r(-1)^r x^{60-7 r}\end{aligned}$
<br/><br/>$\begin{aligned} \therefore 60-7 r =18 \\\\ \Rightarrow 7 r =42 \\\\ \Rightarrow r =6\end{aligned}$
<br/><br/>$\therefore$... | integer | jee-main-2023-online-6th-april-morning-shift | 5,178 |
lsbl0a9f | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | ${ }^{n-1} C_r=\left(k^2-8\right){ }^n C_{r+1}$ if and only if : | [{"identifier": "A", "content": "$2 \\sqrt{2}<\\mathrm{k}<2 \\sqrt{3}$"}, {"identifier": "B", "content": "$2 \\sqrt{2}<\\mathrm{k} \\leq 3$"}, {"identifier": "C", "content": "$2 \\sqrt{3}<\\mathrm{k}<3 \\sqrt{3}$"}, {"identifier": "D", "content": "$2 \\sqrt{3}<\\mathrm{k} \\leq 3 \\sqrt{2}$"}] | ["B"] | null | <p>$${ }^{\mathrm{n}-1} \mathrm{C}_{\mathrm{r}}=(\mathrm{k}^2-8){ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}+1}$$</p>
<p>$$\underbrace{\mathrm{r}+1 \geq 0, \quad \mathrm{r} \geq 0}_{\mathrm{r} \geq 0}$$</p>
<p>$$\begin{aligned}
& \frac{{ }^{n-1} C_r}{{ }^n C_{r+1}}=k^2-8 \\
& \frac{r+1}{n}=k^2-8 \\
& \Rightarrow k^2-8>0 \\
... | mcq | jee-main-2024-online-27th-january-morning-shift | 5,179 |
lsbl8y41 | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | If A denotes the sum of all the coefficients in the expansion of $\left(1-3 x+10 x^2\right)^{\mathrm{n}}$ and B denotes the sum of all the coefficients in the expansion of $\left(1+x^2\right)^n$, then : | [{"identifier": "A", "content": "$\\mathrm{B}=\\mathrm{A}^3$"}, {"identifier": "B", "content": "$3 \\mathrm{A}=\\mathrm{B}$"}, {"identifier": "C", "content": "$A=3 B$"}, {"identifier": "D", "content": "$\\mathrm{A}=\\mathrm{B}^3$"}] | ["D"] | null | <p>Sum of coefficients in the expansion of $$\left(1-3 \mathrm{x}+10 \mathrm{x}^2\right)^{\mathrm{n}}=\mathrm{A}$$</p>
<p>then $$A=(1-3+10)^n=8^n$$ (put $$x=1$$)<?p>
<p>and sum of coefficients in the expansion of</p>
<p>$$\begin{aligned}
& \left(1+x^2\right)^n=B \\
& \text { then } B=(1+1)^n=2^n \\
& A=B^3
\end{aligned... | mcq | jee-main-2024-online-27th-january-morning-shift | 5,180 |
jaoe38c1lsd577xu | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | <p>Let the coefficient of $$x^r$$ in the expansion of $$(x+3)^{n-1}+(x+3)^{n-2}(x+2)+(x+3)^{n-3}(x+2)^2+\ldots \ldots \ldots .+(x+2)^{n-1}$$ be $$\alpha_r$$. If $$\sum_\limits{r=0}^n \alpha_r=\beta^n-\gamma^n, \beta, \gamma \in \mathbb{N}$$, then the value of $$\beta^2+\gamma^2$$ equals _________.</p> | [] | null | 25 | <p>$$\begin{aligned}
& (x+3)^{n-1}+(x+3)^{n-2}(x+2)+(x+3)^{n-3} \\
& (x+2)^2+\ldots \ldots .+(x+2)^{n-1} \\
& \sum \alpha_r=4^{n-1}+4^{n-2} \times 3+4^{n-3} \times 3^2 \ldots \ldots+3^{n-1} \\
& =4^{n-1}\left[1+\frac{3}{4}+\left(\frac{3}{4}\right)^2 \ldots .+\left(\frac{3}{4}\right)^{n-1}\right] \\
& =4^{n-1} \times \f... | integer | jee-main-2024-online-31st-january-evening-shift | 5,181 |
jaoe38c1lse5rrmq | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | <p>In the expansion of $$(1+x)\left(1-x^2\right)\left(1+\frac{3}{x}+\frac{3}{x^2}+\frac{1}{x^3}\right)^5, x \neq 0$$, the sum of the coefficients of $x^3$ and $$x^{-13}$$ is equal to __________.</p> | [] | null | 118 | <p>$$\begin{aligned}
& (1+x)\left(1-x^2\right)\left(1+\frac{3}{x}+\frac{3}{x^2}+\frac{1}{x^3}\right)^5 \\
& =(1+x)\left(1-x^2\right)\left(\left(1+\frac{1}{x}\right)^3\right)^5 \\
& =\frac{(1+x)^2(1-x)(1+x)^{15}}{x^{15}} \\
& =\frac{(1+x)^{17}-x(1+x)^{17}}{x^{15}}
\end{aligned}$$</p>
<p>$$=\operatorname{coeff}\left(\mat... | integer | jee-main-2024-online-31st-january-morning-shift | 5,182 |
1lsg4gw8t | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | <p>Suppose $$2-p, p, 2-\alpha, \alpha$$ are the coefficients of four consecutive terms in the expansion of $$(1+x)^n$$. Then the value of $$p^2-\alpha^2+6 \alpha+2 p$$ equals</p> | [{"identifier": "A", "content": "8"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "6"}, {"identifier": "D", "content": "10"}] | [] | null | <p>$$2-p, p, 2-\alpha, \alpha$$</p>
<p>Binomial coefficients are</p>
<p>$$\begin{aligned}
& { }^n C_r,{ }^n C_{r+1},{ }^n C_{r+2},{ }^n C_{r+3} \text { respectively } \\
\Rightarrow \quad & { }^n C_r+{ }^n C_{r+1}=2 \\
\Rightarrow \quad & { }^{n+1} C_{r+1}=2 \quad \ldots . .(1)
\end{aligned}$$</p>
<p>Also, $${ }^{\math... | mcq | jee-main-2024-online-30th-january-evening-shift | 5,184 |
1lsg5bloa | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | <p>Let $$\alpha=\sum_\limits{k=0}^n\left(\frac{\left({ }^n C_k\right)^2}{k+1}\right)$$ and $$\beta=\sum_\limits{k=0}^{n-1}\left(\frac{{ }^n C_k{ }^n C_{k+1}}{k+2}\right)$$ If $$5 \alpha=6 \beta$$, then $$n$$ equals _______.</p> | [] | null | 10 | <p>$$\begin{aligned}
\alpha= & \sum_{k=0}^n \frac{{ }^n C_k \cdot{ }^n C_k}{k+1} \cdot \frac{n+1}{n+1} \\
& =\frac{1}{n+1} \sum_{k=0}^n{ }^{n+1} C_{k+1} \cdot{ }^n C_{n-k} \\
\alpha & =\frac{1}{n+1} \cdot{ }^{2 n+1} C_{n+1} \\
\beta & =\sum_{k=0}^{n-1} C_k \cdot \frac{{ }^n C_{k+1}}{k+2} \frac{n+1}{n+1} \\
& \frac{1}{n... | integer | jee-main-2024-online-30th-january-evening-shift | 5,185 |
luxwdx5b | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | <p>The sum of the coefficient of $$x^{2 / 3}$$ and $$x^{-2 / 5}$$ in the binomial expansion of $$\left(x^{2 / 3}+\frac{1}{2} x^{-2 / 5}\right)^9$$ is</p> | [{"identifier": "A", "content": "19/4"}, {"identifier": "B", "content": "69/16"}, {"identifier": "C", "content": "63/16"}, {"identifier": "D", "content": "21/4"}] | ["D"] | null | <p>$$
\begin{aligned}
& T_{r+1}={ }^9 C_r\left(\frac{x^{-2 / 5}}{2}\right)^r\left(x^{2 / 3}\right)^{9-r} \\
& ={ }^9 C_r \frac{1}{2^r} x^{\frac{2}{3}(9-r)+\left(\frac{-2 r}{5}\right)} \\
& ={ }^9 C_r \cdot \frac{1}{2^r} \cdot x^{6-\frac{16 r}{15}}
\end{aligned}
$$</p>
<p>For coefficient of $$x^{2 / 3}$$</p>
<p>$$\begin... | mcq | jee-main-2024-online-9th-april-evening-shift | 5,186 |
luy6z5lw | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | <p>The coefficient of $$x^{70}$$ in $$x^2(1+x)^{98}+x^3(1+x)^{97}+x^4(1+x)^{96}+\ldots+x^{54}(1+x)^{46}$$ is $${ }^{99} \mathrm{C}_{\mathrm{p}}-{ }^{46} \mathrm{C}_{\mathrm{q}}$$. Then a possible value of $$\mathrm{p}+\mathrm{q}$$ is :</p> | [{"identifier": "A", "content": "61"}, {"identifier": "B", "content": "83"}, {"identifier": "C", "content": "55"}, {"identifier": "D", "content": "68"}] | ["B"] | null | <p>$$x^2(1+x)^{98}+x^3(1+x)^{97}+\ldots+x^{54}(1+x)^{46}$$</p>
<p>It is a G.P. with first term $$=x^2(1+x)^{98}$$</p>
<p>and common ratio $$=\frac{x}{1+x}$$</p>
<p>sum of these term $$=x^2(1+x)^{98}\left(\frac{\left(\frac{x}{1+x}\right)^{53}-1}{\frac{x}{1+x}-1}\right)$$</p>
<p>$$=x^2(1+x)^{98}\left((1+x)-x^{53}(1+x)^{-... | mcq | jee-main-2024-online-9th-april-morning-shift | 5,187 |
lv0vxdwu | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | <p>Let $$a=1+\frac{{ }^2 \mathrm{C}_2}{3 !}+\frac{{ }^3 \mathrm{C}_2}{4 !}+\frac{{ }^4 \mathrm{C}_2}{5 !}+...., \mathrm{b}=1+\frac{{ }^1 \mathrm{C}_0+{ }^1 \mathrm{C}_1}{1 !}+\frac{{ }^2 \mathrm{C}_0+{ }^2 \mathrm{C}_1+{ }^2 \mathrm{C}_2}{2 !}+\frac{{ }^3 \mathrm{C}_0+{ }^3 \mathrm{C}_1+{ }^3 \mathrm{C}_2+{ }^3 \mathrm... | [] | null | 8 | <p>$$\begin{aligned}
& a=1+\frac{{ }^2 C_2}{3!}+\frac{{ }^3 C_2}{4!}+\frac{{ }^4 C_2}{5!}+\ldots \\
& b=1+\frac{{ }^1 C_0+{ }^1 C_1}{1!}+\frac{{ }^2 C_0+{ }^2 C_1+{ }^2 C_2}{2!}+\ldots \\
& b=1+\frac{2}{1!}+\frac{2^2}{2!}+\frac{2}{3!}+\ldots=e^2
\end{aligned}$$</p>
<p>Using $$e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x}{... | integer | jee-main-2024-online-4th-april-morning-shift | 5,188 |
lv2eqvbz | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | <p>If the coefficients of $$x^4, x^5$$ and $$x^6$$ in the expansion of $$(1+x)^n$$ are in the arithmetic progression, then the maximum value of $$n$$ is:</p> | [{"identifier": "A", "content": "28"}, {"identifier": "B", "content": "21"}, {"identifier": "C", "content": "7"}, {"identifier": "D", "content": "14"}] | ["D"] | null | <p>$$\begin{aligned}
& (1+x)^n={ }^n C_0+{ }^n C_1 x^1+{ }^n C_2 x^2+\ldots{ }^n C_n x^n \\
& { }^n C_4,{ }^n C_5 \&{ }^n C_6 \text { are in A.P. } \\
& { }^n C_5-{ }^n C_4={ }^n C_6-{ }^n C_5 \\
& \Rightarrow \frac{n!}{5!(n-5)!}-\frac{n!}{4!(n-4)!}=\frac{n!}{6!(n-6)!}-\frac{n!}{5!(n-5)!} \\
& \Rightarrow 30(n-9)(n-6)=... | mcq | jee-main-2024-online-4th-april-evening-shift | 5,189 |
xQSkrUiNigxHOp8P | maths | circle | basic-theorems-of-a-circle | The equation of a circle with origin as a center and passing through an equilateral triangle whose median is of length $$3$$$$a$$ is : | [{"identifier": "A", "content": "$${x^2}\\, + \\,{y^2} = 9{a^2}$$ "}, {"identifier": "B", "content": "$${x^2}\\, + \\,{y^2} = 16{a^2}$$"}, {"identifier": "C", "content": "$${x^2}\\, + \\,{y^2} = 4{a^2}$$ "}, {"identifier": "D", "content": "$${x^2}\\, + \\,{y^2} = {a^2}$$"}] | ["C"] | null | Let $$ABC$$ be an equilateral triangle, whose median is $$AD.$$
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267493/exam_images/qh2bpsmsygtrhhpn1feb.webp" loading="lazy" alt="AIEEE 2002 Mathematics - Circle Question 157 English Explanation">
<br><br>Given $$AD=3a.$$
<b... | mcq | aieee-2002 | 5,190 |
erkEpvhy5MhZbgZh | maths | circle | basic-theorems-of-a-circle | The point diametrically opposite to the point $$P(1, 0)$$ on the circle $${x^2} + {y^2} + 2x + 4y - 3 = 0$$ is : | [{"identifier": "A", "content": "$$(3, -4)$$"}, {"identifier": "B", "content": "$$(-3, 4)$$ "}, {"identifier": "C", "content": "$$(-3, -4)$$"}, {"identifier": "D", "content": "$$(3, 4)$$"}] | ["C"] | null | The given circle is $${x^2} + {y^2} + 2x + 4y - 3 = 0$$
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266849/exam_images/earxtruibjeyrajhopbc.webp" loading="lazy" alt="AIEEE 2008 Mathematics - Circle Question 144 English Explanation">
<br><br>Center $$(-1,-2)$$
<br><br... | mcq | aieee-2008 | 5,191 |
ExtzD08AoBipFNWA | maths | circle | basic-theorems-of-a-circle | Three distinct points A, B and C are given in the 2 -dimensional coordinates plane such that the ratio of the distance of any one of them from the point $$(1, 0)$$ to the distance from the point $$(-1, 0)$$ is equal to $${1 \over 3}$$. Then the circumcentre of the triangle ABC is at the point : | [{"identifier": "A", "content": "$$\\left( {{5 \\over 4},0} \\right)$$ "}, {"identifier": "B", "content": "$$\\left( {{5 \\over 2},0} \\right)$$"}, {"identifier": "C", "content": "$$\\left( {{5 \\over 3},0} \\right)$$"}, {"identifier": "D", "content": "$$\\left( {0,0} \\right)$$ "}] | ["A"] | null | Given that
<br><br>$$P\left( {1,0} \right),Q\left( { - 1,0} \right)$$
<br><br>and $${{AP} \over {AQ}} = {{BP} \over {BQ}} = {{CP} \over {CQ}} = {1 \over 3}$$
<br><br>$$ \Rightarrow 3AP = AQ$$
<br><br>$$\,\,\,\,\,\,$$ Let $$A = (x,y)$$ then $$3AP = AQ \Rightarrow 9A{P^2} = A{Q^2}$$
<br><br>$$ \Rightarrow 9{\left( {x -... | mcq | aieee-2009 | 5,192 |
53q58LReyCXabFV0 | maths | circle | basic-theorems-of-a-circle | Locus of the image of the point $$(2, 3)$$ in the line $$\left( {2x - 3y + 4} \right) + k\left( {x - 2y + 3} \right) = 0,\,k \in R,$$ is a : | [{"identifier": "A", "content": "circle of radius $$\\sqrt 2 $$."}, {"identifier": "B", "content": "circle of radius $$\\sqrt 3 $$."}, {"identifier": "C", "content": "straight line parallel to $$x$$-axis "}, {"identifier": "D", "content": "straight line parallel to $$y$$-axis "}] | ["A"] | null | Intersection point of $$2x - 3y + 4 = 0$$
<br><br>and $$x-2y+3=0$$ is $$(1, 2)$$
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l91npsqa/a856b112-8560-4eba-878d-d29d808c6c0a/1868e920-47fd-11ed-8757-0f869593f41f/file-1l91npsqb.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/i... | mcq | jee-main-2015-offline | 5,193 |
EfmKOioHmsuvP1wBRvjgy2xukf0zcypw | maths | circle | basic-theorems-of-a-circle | The diameter of the circle, whose centre lies on
the line x + y = 2 in the first quadrant and which
touches both the lines x = 3 and y = 2, is
_______ . | [] | null | 3 | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263616/exam_images/ssi5vdlnvmrlmfb1iuud.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 3rd September Morning Slot Mathematics - Circle Question 96 English Explanation">
<br>$$ \because ... | integer | jee-main-2020-online-3rd-september-morning-slot | 5,196 |
xYtfGaltEQHT5L76yWjgy2xukfal1q5d | maths | circle | basic-theorems-of-a-circle | Let PQ be a diameter of the circle x<sup>2</sup> + y<sup>2</sup> = 9. If $$\alpha $$ and $$\beta $$ are the lengths of the perpendiculars from P and Q on the straight line,<br/> x + y = 2 respectively, then the maximum value of $$\alpha\beta $$ is _____. | [] | null | 7 | Let $$P(3\cos \theta ,\,3\sin \theta )$$<br><br>$$Q( - 3\cos \theta ,\, - 3\sin \theta )$$<br><br>$$\alpha = \left| {{{3\cos \theta + 3\sin \theta - 2} \over {\sqrt 2 }}} \right|$$<br><br>$$\beta = \left| {{{ - 3\cos \theta - 3\sin \theta - 2} \over {\sqrt 2 }}} \right|$$<br><br>$$\alpha \beta = \left| {{{{{\lef... | integer | jee-main-2020-online-4th-september-evening-slot | 5,197 |
0m8EX9yisbE45d5O431klrmflld | maths | circle | basic-theorems-of-a-circle | Let a point P be such that its distance from the point (5, 0) is thrice the distance of P from the point ($$-$$5, 0). If the locus of the point P is a circle of radius r, then 4r<sup>2</sup> is equal to ________ | [] | null | 56 | Let P(h, k)<br><br>Given<br><br>PA = 3PB<br><br>PA<sup>2</sup> = 9PB<sup>2</sup><br><br>$$ \Rightarrow $$ (h $$-$$ 5)<sup>2</sup> + k<sup>2</sup> = 9[(h + 5)<sup>2</sup> + k<sup>2</sup>]<br><br>$$ \Rightarrow $$ 8h<sup>2</sup> + 8k<sup>2</sup> + 100h + 200 = 0<br><br>$$ \therefore $$ Locus<br><br>$${x^2} + {y^2} + \lef... | integer | jee-main-2021-online-24th-february-evening-slot | 5,198 |
1ks09eg6q | maths | circle | basic-theorems-of-a-circle | Let P and Q be two distinct points on a circle which has center at C(2, 3) and which passes through origin O. If OC is perpendicular to both the line segments CP and CQ, then the set {P, Q} is equal to : | [{"identifier": "A", "content": "{(4, 0), (0, 6)}"}, {"identifier": "B", "content": "$$\\{ (2 + 2\\sqrt 2 ,3 - \\sqrt 5 ),(2 - 2\\sqrt 2 ,3 + \\sqrt 5 )\\} $$"}, {"identifier": "C", "content": "$$\\{ (2 + 2\\sqrt 2 ,3 + \\sqrt 5 ),(2 - 2\\sqrt 2 ,3 - \\sqrt 5 )\\} $$"}, {"identifier": "D", "content": "{($$-$$1, 5), (5,... | ["D"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265883/exam_images/j4eb4u3olf4aobpzkvfj.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263397/exam_images/a0quemgl8enbtkiviozm.webp"><img src="https://res.c... | mcq | jee-main-2021-online-27th-july-morning-shift | 5,200 |
1ks0au11w | maths | circle | basic-theorems-of-a-circle | Let $$A = \{ (x,y) \in R \times R|2{x^2} + 2{y^2} - 2x - 2y = 1\} $$, $$B = \{ (x,y) \in R \times R|4{x^2} + 4{y^2} - 16y + 7 = 0\} $$ and $$C = \{ (x,y) \in R \times R|{x^2} + {y^2} - 4x - 2y + 5 \le {r^2}\} $$.<br/><br/>Then the minimum value of |r| such that $$A \cup B \subseteq C$$ is equal to | [{"identifier": "A", "content": "$${{3 + \\sqrt {10} } \\over 2}$$"}, {"identifier": "B", "content": "$${{2 + \\sqrt {10} } \\over 2}$$"}, {"identifier": "C", "content": "$${{3 + 2\\sqrt 5 } \\over 2}$$"}, {"identifier": "D", "content": "$$1 + \\sqrt 5 $$"}] | ["C"] | null | $${S_1}:{x^2} + {y^2} - x - y - {1 \over 2} = 0;{C_1}\left( {{1 \over 2},{1 \over 2}} \right)$$<br><br>$${r_1} = \sqrt {{1 \over 4} + {1 \over 4} + {1 \over 2}} = 1$$<br><br>$${S_2}:{x^2} + {y^2} - 4y + {7 \over 4} = 0;{C_2}:(0,2)$$<br><br>$${r_2} = \sqrt {4 - {7 \over 4}} = {3 \over 2}$$<br><br>$${S_3} = {x^2} + {y^... | mcq | jee-main-2021-online-27th-july-morning-shift | 5,201 |
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