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values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
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Example 1 Given $a^{2}+b^{2}-k a b=1$, $c^{2}+d^{2}-k c d=1$, $a, b, c, d$, $k \in \mathbf{R}$, and $|k|<2$, prove that
$$|a c-b d| \leqslant \frac{2}{\sqrt{4-k^{2}}}$$
Equality in (1) holds if and only if $\frac{2-k}{2+k}=\frac{(a-b)(c-d)}{(a+b)(c+d)}$, i.e., $k=\frac{b c+a d}{a c+b d}$. | Proof: Let the parameter $t \neq 0$, and applying the Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
4|a c-b d|^{2}=|(a+b)(c-d)+(a-b)(c+d)|^{2}= \\
\left|\frac{t(a+b) \cdot(c-d)}{t}+(a-b)(c+d)\right|^{2} \leqslant \\
{\left[t^{2}(a+b)^{2}+(a-b)^{2}\right]\left[\frac{(c-d)^{2}}{t^{2}}+(c+d)^{2}\right]=} \\
{\left... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,723 |
Example 2 (Self-created problem, 1989. 04. 06) $a, b, c$ are known non-zero real numbers, $x, y, z, w \in \mathbf{R}$, then
$$\frac{a x y+b y z+c z w}{x^{2}+y^{2}+z^{2}+w^{2}} \leqslant \frac{1}{2}\left[\sqrt{(a+c)^{2}+b^{2}}+\sqrt{(a-c)^{2}+b^{2}}\right]$$
Equality holds if and only if $\frac{y}{x}=a \alpha^{2}, \fra... | Let $\alpha, \beta$ be two non-zero parameters, then
$$\begin{aligned}
2 a x y+2 b y z+2 c z w= & {\left[(a \alpha x)^{2}+\left(\frac{y}{\alpha}\right)^{2}-\left(a \alpha x-\frac{y}{\alpha}\right)^{2}\right]+} \\
& {\left[(b \beta y)^{2}+\left(\frac{z}{\beta}\right)^{2}-\left(b \beta y-\frac{z}{\beta}\right)^{2}\right]... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,724 |
$1 \cdot 1$ Let $S(n)$ denote the sum of all digits of the natural number $n$.
（1）Does there exist a natural number $n$ such that $n+S(n)=1980$?
（2）Prove: Among any two consecutive natural numbers, one can be expressed in the form $n+S(n)$, where $n$ is some natural number. | [Solution] (1) $1962+S(1962)=1980$.
(2) We denote $S(n)+n$ as $S_{n}$. If the last digit of the number $n$ is 9, then $S_{n+1}2$. Select the largest $N$ such that $S_{N}<m$, then $S_{N+1} \geqslant m$. Clearly, the last digit of $N$ is not 9. Therefore, $S_{N+1}=m$ or $S_{N+1}=m+1$. It is also clear that $S_{1}=2$, thu... | 1962+S(1962)=1980 | Number Theory | proof | Yes | Yes | inequalities | false | 734,725 |
$1 \cdot 2$ Proof: For any natural number $k$, there exist infinitely many natural numbers $t$ (in decimal notation) that do not contain the digit 0, such that the sum of the digits of $t$ and $kt$ are the same. | [Proof] If the number $k$ is denoted as
$$k=\overline{a_{n} a_{n-1} \cdots a_{0}},$$
let's assume $a_{0} \geqslant 1, t$ represents the number composed of $m$ nines,
$$t=\underset{m \uparrow}{99 \cdots 9}=10^{m}-1$$
where $m>n$. Then,
$$k t=a_{n} a_{n-1} \cdots\left(a_{0}-1\right) 99 \cdots 9\left(9-a_{n}\right)\left... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 734,726 |
1- 11 Integers $a$, $b$, $c$ form a geometric sequence, in the decimal number $N=a^{3}+b^{3}+c^{3}-$ $3abc$, the last digit is 0, and the second last digit is 2, is this possible? | ［Solution］ If the last two digits of a decimal number $N$ are 20, then
$N=100 M+20$ (where $M$ is an integer)
it is divisible by 5 but not by 25.
On the other hand, we can prove that if $N$ is divisible by a prime number $p$, then it is divisible by $p^{2}$.
In fact, because the common ratio $q$ of the sequence is a ra... | proof | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 734,727 |
1- 104 Natural numbers $p$ and $q$ are coprime, and the interval $[0,1]$ is divided into $p+q$ equal intervals. Prove: except for the leftmost and rightmost intervals, each of the remaining intervals contains one of the following $p+q-2$ numbers:
$$\frac{1}{p}, \frac{2}{p}, \cdots, \frac{p-1}{p}, \frac{1}{q}, \frac{2}{... | [Proof] Since $p$ and $q$ are coprime, each of them is coprime with $n=p+q$. Therefore, $\frac{i}{p}, \frac{j}{q}, \frac{i+j}{n} (i=1,2, \cdots, p-1 ; j=1,2, \cdots, q-1)$ are all distinct. Moreover, $\frac{i+j}{p+q}$ is always between $\frac{i}{p}$ and $\frac{j}{q}$. Thus, all fractions $\frac{i}{p}$ and $\frac{j}{q}$... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 734,728 |
7.4 Arrange $\frac{n(n+1)}{2}$ different numbers randomly into a triangular array as shown in the figure. Let $M_{k}$ be the maximum number in the $k$-th row (from top to bottom). Try to find the probability that $M_{1}<M_{2}<\cdots<M_{n}$. | [Solution] Let the required probability be $p_{n}$, then $p_{1}=1$.
For a table with $n+1$ rows, the largest number $\frac{(n+1)(n+2)}{2}$ should be in the $n+1$-th row, the probability of which is
$$\frac{n+1}{\frac{(n+1)(n+2)}{2}}=\frac{2}{n+2} \text {. }$$
Then the $(n+1)$-th row can be filled arbitrarily, and fina... | \frac{2^{n}}{(n+1)!} | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 734,729 |
7. 7 Given six points $A, B, C, D, E, F$ chosen randomly on a given circumference, these points are chosen independently and are equally likely with respect to arc length. Find the probability that triangle $A B C$ and triangle $D E F$ do not intersect (i.e., have no common points). | [Solution] Note that the number of cyclic permutations of six distinct points distributed on a circle is
$$p_{5}^{5}=5!=120$$
And due to the symmetry of the distribution, these permutations have the same probability. The number of different permutations where triangle $A B C$ and triangle $D E F$ do not intersect is
$... | \frac{3}{10} | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 734,730 |
7. 8 If a positive divisor of $10^{99}$ is chosen at random, the probability that it is exactly a multiple of $10^{88}$ is $\frac{m}{n}$, where $m$ and $n$ are coprime, find $m+n$. | ［Solution］The divisors of $10^{99}$ all have the form
$$2^{a} \cdot 5^{b}$$
where $a$ and $b$ satisfy $0 \leqslant a \leqslant 99,0 \leqslant b \leqslant 99$.
Therefore, $10^{99}$ has
$$(99+1)(99+1)=100 \times 100$$
positive divisors.
Among these positive divisors, to be a multiple of $10^{88}=2^{88} \cdot 5^{88}$, i... | 634 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 734,731 |
7.9 A gardener is going to plant three maple trees, four oak trees, and five birch trees in a row. He randomly determines the order of these trees, and all different arrangements are equally probable. Use $\frac{m}{n}$ to represent the probability that no two birch trees are adjacent (simplified to the lowest terms), a... | [Solution] There are 12! different arrangements of 12 trees.
There are 7 non-birch trees (3 maple trees and 4 oak trees). To ensure that no two birch trees are adjacent, the 5 birch trees can be placed in the gaps between the 7 non-birch trees. Since there are 8 gaps among the 7 non-birch trees, the number of ways to a... | 106 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 734,732 |
7・10 A fishing boat is fishing in the territorial sea of a foreign country without permission. Each time it casts a net, it causes the same value of loss to the country's fishing yield. The probability of the boat being detained by the foreign coast patrol during each net casting is equal to $1 / k$, where $k$ is a nat... | [Solution] The probability that the fishing boat is not detained during the first net casting is $1-\frac{1}{k}$. Since the events of being detained or not during each net casting are independent, the probability of not being detained after casting the net $n$ times is $\left(1-\frac{1}{k}\right)^{n}$. Therefore, the e... | n=k-1 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 734,733 |
7. 12 For a known sequence of distinct real numbers $r_{1}, r_{2}, r_{3}, \cdots$, $r_{n}$, one operation consists of comparing the second term with the first term, and swapping them if and only if the second term is smaller; this process continues until the last term is compared with its new preceding term, and they a... | [Solution] We note the following fact:
For any sequence
$$r_{1}, r_{2}, \cdots, r_{k}$$
after performing the operation defined in the problem once, the last number must be the largest number in the sequence.
Therefore, in an initial sequence
$r_{1}, r_{2}, \cdots, r_{20}, \cdots, r_{30}, r_{31}, \cdots, r_{40}$, for $... | 931 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 734,735 |
7・ 13 A biologist wants to calculate the number of fish in a lake; on May 1, he randomly caught 60 fish and marked them, then released them back into the lake. On September 1, he randomly caught 70 fish again and found that 3 of them were marked. He assumes that 25% of the fish in the lake on May 1 were no longer in th... | [Solution] The 60 fish marked on May 1, by September 1, possibly remain in the lake as
$$60(1-25 \%)=45 \text { (fish). }$$
Of these 45 fish, 3 marked fish were caught, so the proportion of marked fish caught out of all marked fish is $\frac{3}{45}=\frac{1}{15}$.
These three marked fish were found among the 70 fish c... | 840 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 734,736 |
1. 105 Let $n$ be an odd number. Prove that there exist $2n$ integers $a_{1}, a_{2}, \cdots, a_{n}$ and $b_{1}, b_{2}$, $\cdots, b_{n}$, such that for any integer $k(0<k<n)$, the following $3n$ numbers
$$a_{i}+a_{i+1}, a_{i}+b_{i}, b_{i}+b_{i+k}, i=1,2, \cdots, n$$
when divided by $3n$ yield distinct remainders, where... | [Solution] Let
$a_{i}=3 i-2, b_{i}=3 i-3, i=1,2, \cdots, n$, then these $2 n$ numbers satisfy the requirements of the problem.
In fact, because
$$\begin{aligned}
d_{i} & =a_{i}+a_{i+1}=6 i-1, i=1,2, \cdots, n . \\
d_{j} & -d_{i}=6(j-i), 1 \leqslant i<j \leqslant n
\end{aligned}$$
it follows that $d_{j}-d_{i}<6 n$, an... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 734,737 |
7-15 In a given regular $(2n+1)$-sided polygon, three different vertices are randomly selected. If all such selections are equally likely. Find the probability that the center of the regular polygon lies inside the triangle formed by the randomly chosen three points. | [Solution] First, fix one vertex of the triangle, denoted as $p_{0}$, and the vertices on both sides of $p_{0}$ are sequentially denoted as $p_{1}, p_{2}, \cdots, p_{n}$ and $p_{-1}, p_{-2}, \cdots, p_{-n}$.
For the center of the regular polygon to be inside the triangle, it is necessary and sufficient that the other ... | \frac{n+1}{2(2 n-1)} | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 734,738 |
7・17 Father, mother, and son decide to hold a family competition of some game, with each match involving two participants, and there are no draws. Since the father is the weakest player, he gets to choose the two participants for the first match. The winner of each match plays against the person who did not participate... | [Proof] Let $F$, $M$, and $S$ represent the father, mother, and son, respectively. Use the notation $W > L$ to indicate that $W$ wins over $L$ in a match.
If $F$ and $M$ play the first match, then for any of the following three mutually independent cases, $F$ can always win the championship:
(1) $F > M, F > S$;
(2) $F... | proof | Logic and Puzzles | proof | Yes | Yes | inequalities | false | 734,739 |
7・18 Given three congruent $n$-sided dice, their corresponding faces are labeled with the same arbitrary integers. Prove that if they are randomly rolled, the probability that the sum of the numbers on the three upward faces is divisible by 3 is greater than or equal to $\frac{1}{4}$. | [Proof] Let a die with $n$ faces have $n_{0}$ faces with integers divisible by 3, $n_{1}$ faces with integers that leave a remainder of 1 when divided by 3, and $n_{2}$ faces with integers that leave a remainder of 2 when divided by 3. Here,
$$\begin{array}{l}
n_{0} \geqslant 0, n_{1} \geqslant 0, n_{2} \geqslant 0 \\
... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 734,740 |
7. 19 Salmon moving along the mountain stream must pass through two waterfalls. In this experiment, the probability of a salmon passing the first waterfall is $p>0$, and the probability of passing the second waterfall is $q>0$. Assume that the trials of passing the waterfalls are independent. Find the probability that,... | [Solution] Let $A_{n}$ be the event that the salmon fails to leap the first waterfall in $n$ trials, and $B_{n}$ be the event that the salmon fails to leap both waterfalls in $n$ trials. Since the probability of failing to leap the first waterfall in one trial is $1-p$, and the trials are independent, we have
$$p\left(... | \frac{(p-q)\left(\frac{1-p}{1-q}\right)^{n}}{p-q\left(\frac{1-p}{1-q}\right)^{n}} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 734,741 |
7.20 A single-player card game has the following rules: Place 6 pairs of different cards into a backpack. The player draws cards randomly from the backpack and returns them, but when a pair is drawn, it is set aside. If the player always draws three cards at a time, and if the three cards drawn are all different (i.e.,... | [Solution] Let there be $n$ pairs of distinct cards in the bag, where $n \geqslant 2$. Then the probability that 2 out of the first 3 cards drawn form a pair is
the number of ways to draw 3 cards with 2 forming a pair
the number of ways to draw 3 cards
$$\begin{array}{l}
=\frac{C_{n}^{1} \cdot C_{2 n-2}^{1}}{C_{2 n}^{... | 394 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 734,742 |
$8 \cdot 1$ Arrange all powers of 3 and the sums of any finite number of distinct powers of 3 in an increasing sequence:
$$1,3,4,9,10,12,13, \cdots$$
Find the 100th term of this sequence. | [Solution] Let the $n$-th term of the sequence be denoted as $a_{n}$. Clearly, $a_{n}$ is a positive integer, and its ternary representation uses only the digits 0 and 1. Conversely, any natural number whose ternary representation uses only the digits 0 and 1 must be a term in this sequence. Therefore, the binary repre... | 981 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 734,743 |
$$\begin{array}{ll}
8 \cdot 3 & \text { Let the sequence } x_{1}, x_{2}, x_{3}, \cdots \text { satisfy } \\
& 3 x_{n}-x_{n-1}=n, n=2,3, \cdots \\
\text { and } & \left|x_{1}\right|<1971 .
\end{array}$$
Find \( x_{1971} \), accurate to 0.000001. | [Solution] Clearly, the sequence $y_{n}=\frac{1}{2} n-\frac{1}{4}, n=1,2,3, \cdots$ satisfies
$$3 y_{n}-y_{n-1}=n, n=2,3, \cdots$$
Therefore, $\left|y_{n}-x_{n}\right|=\frac{1}{3}\left|y_{n-1}-x_{n-1}\right|$,
from which we can deduce that $\left|y_{1971}-x_{1971}\right|=\left(\frac{1}{3}\right)^{1970}\left|y_{1}-x_{1... | 985.250000 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 734,744 |
1. The MO brand football is made of several polygonal leather pieces sewn together with threads of three different colors, and it has the following characteristics:
(1) Any side of a polygonal leather piece is exactly sewn to a side of equal length of another polygonal leather piece using a thread of one color;
(2) At ... | [Proof] Let the three colors be red, yellow, and blue.
We divide all nodes into two categories. If, when moving counterclockwise around a node, the colors of the three lines passing through the node are red, blue, and yellow in sequence, we call this node a first-class node. If the sequence is red, yellow, and blue, we... | proof | Combinatorics | proof | Yes | Yes | inequalities | false | 734,745 |
8-4 Let the sequence of positive integers $\left\{a_{n}\right\}$ satisfy
$$a_{n+3}=a_{n+2}\left(a_{n+1}+2 a_{n}\right), n=1,2, \cdots$$
and $a_{6}=2288$. Find $a_{1}, a_{2}, a_{3}$. | ［Solution］From the recursive relation, we get
$$\begin{aligned}
a_{6} & =a_{5}\left(a_{4}+2 a_{3}\right)=a_{4}\left(a_{3}+2 a_{2}\right)\left(a_{4}+2 a_{3}\right) \\
& =a_{3}\left(a_{2}+2 a_{1}\right)\left(a_{3}+2 a_{2}\right)\left[a_{3}\left(a_{2}+2 a_{1}\right)+2 a_{3}\right] \\
& =a_{3}^{2}\left(a_{3}+2 a_{2}\right)... | a_{1}=5, a_{2}=1, a_{3}=2 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 734,746 |
$8 \cdot 6$ Let the sequence $a_{0}, a_{1}, a_{2}, \cdots$ satisfy
$$a_{0}=a_{1}=11, a_{m+n}=\frac{1}{2}\left(a_{2 m}+a_{2 n}\right)-(m-n)^{2}, m, n \geqslant 0 .$$
Find $a_{45}$. | [Solution] From $a_{1}=a_{1+0}=\frac{1}{2}\left(a_{2}+a_{0}\right)-1$, we have $a_{2}=13$.
For $n \geqslant 1$, by the assumption,
$$a_{2 n}=a_{n+1+n-1}=\frac{1}{2}\left(a_{2 n+2}+a_{2 n-2}\right)-4 \text {, }$$
Thus, let $b_{n}=a_{2 n}$, we get
$$b_{0}=11, b_{1}=13 \text {, and } \quad b_{n+1}=2 b_{n}-b_{n-1}+8, n \g... | 1991 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 734,747 |
8.7 Given $a_{1}=1996$,
$$a_{k}=\left[\sqrt{a_{1}+a_{2}+\cdots+a_{k-1}}\right], k=2,3, \cdots$$
where $[x]$ denotes the greatest integer not exceeding $x$. Find $a_{1966}$. | [Solution] $a_{1}=1966=44^{2}+30$, so $a_{2}=a_{3}=44$ and
$$a_{1}+a_{2}+a_{3}=45^{2}+29$$
From this, we know that $a_{4}=a_{5}=45$ and
$$a_{1}+a_{2}+a_{3}+a_{4}+a_{5}=46^{2}+28$$
By analogy, we get $a_{60}=a_{61}=73$ and
$$a_{1}+a_{2}+\cdots+a_{61}=74^{2}$$
Thus, $a_{62}=a_{63}=a_{64}=74$ and
$$a_{1}+a_{2}+\cdots+a... | 1024 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 734,748 |
$8 \cdot 8$ Let $x_{1}, x_{2}$ be two natural numbers less than 1000, and let
$$\begin{array}{l}
x_{3}=\left|x_{1}-x_{2}\right|, x_{4}=\min \left\{\left|x_{1}-x_{2}\right|,\left|x_{1}-x_{3}\right|,\left|x_{2}-x_{3}\right|\right\} \cdots, \\
x_{k}=\min \left\{\left|x_{i}-x_{j}\right| \mid 1 \leqslant i, j \leqslant k-1,... | [Proof] Without loss of generality, let $x_{1} \geqslant x_{2}$. Clearly, $x_{1}, x_{2}, x_{3}, x_{4}, \cdots$ is a sequence of non-negative integers and $x_{1} \geqslant x_{2} \geqslant x_{3} \geqslant x_{4} \geqslant \cdots$. Furthermore, we have
$$x_{k} \leqslant x_{k-2}-x_{k-1}, k=3,4, \cdots$$
Assume $x_{21} \neq... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 734,749 |
$\begin{array}{c}8 \cdot 9 \text { - Let } \quad a_{n}=\frac{1}{(n+1) \sqrt{n}+n \sqrt{n+1}}, n=1,2,3, \cdots \text { Find } \\ a_{1}+a_{2}+\cdots+a_{99} .\end{array}$ | [Solution] Since
$$\begin{aligned}
a_{n} & =\frac{1}{(n+1) \sqrt{n}+n \sqrt{n+1}}=\frac{(n+1) \sqrt{n}-n \sqrt{n+1}}{(n+1)^{2} n-n^{2}(n+1)} \\
& =\frac{(n+1) \sqrt{n}-n \sqrt{n+1}}{n(n+1)}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}
\end{aligned}$$
Therefore, \(a_{1}+a_{2}+\cdots+a_{99}=\sum_{n=1}^{99}\left(\frac{1}{\sqr... | \frac{9}{10} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 734,750 |
8-10 Let the sequence $\left\{a_{n}\right\}$ satisfy
$$a_{n}=a_{n-1}-a_{n-2}, n \geqslant 3 .$$
If the sum of its first 1492 terms is 1985, and the sum of the first 1985 terms is 1492, find the sum of its first 2001 terms. | ［Solution］ For any natural number $n$, by the assumption we have
$$\begin{array}{l}
a_{n}, a_{n+1}, a_{n+2}=a_{n+1}-a_{n}, a_{n+3}=a_{n+2}-a_{n+1}=-a_{n} \\
a_{n+4}=a_{n+3}-a_{n+2}=-a_{n+1}, a_{n+5}=a_{n}-a_{n+1}, a_{n+6}=a_{n}
\end{array}$$
From this, we can also get $a_{n}+a_{n+1}+a_{n+2}+a_{n+3}+a_{n+4}+a_{n+5}=0$.... | 986 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 734,751 |
8-11 $n^{2}(n \geqslant 4)$ positive numbers are arranged in $n$ rows and $n$ columns:
$$\begin{array}{l}
a_{11} a_{12} a_{13} a_{14} \cdots a_{1 n} \\
a_{21} a_{22} a_{23} a_{24} \cdots a_{2 n} \\
a_{31} a_{32} a_{33} a_{34} \cdots a_{3 n} \\
a_{41} a_{42} a_{43} a_{44} \cdots a_{4 n} \\
\cdots \cdots \\
a_{n 1} a_{n ... | [Solution] Since each row of numbers forms an arithmetic sequence, and $a_{42}=\frac{1}{8}, a_{43}=\frac{3}{16}$, the common difference of the fourth row is $\frac{1}{16}$, and thus
$$a_{4 k}=\frac{k}{16}, k=1,2,3,4, \cdots, n$$
Furthermore, since each column of numbers forms a geometric sequence with the same common ... | 2-\frac{n+2}{2^{n}} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 734,752 |
8・13 Let the sequence $\left\{x_{n}\right\}$ satisfy: $x_{1}=\frac{1}{2}$, and
$$x_{k+1}=x_{k}+x_{k}^{2}, k=1,2, \cdots$$
Find the integer part of $\frac{1}{x_{1}+1}+\frac{1}{x_{2}+1}+\cdots+\frac{1}{x_{100}+1}$. | ［Solution］From the recursive relation, it is easy to know
$$01, x_{101}>x_{3}$$ which implies
$$1<\frac{1}{x_{1}}-\frac{1}{x_{101}}<2$$
Thus, the integer part of $\frac{1}{x_{1}+1}+\frac{1}{x_{2}+1}+\cdots+\frac{1}{x_{100}+1}$ is 1. | 1 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 734,754 |
107 Can the numbers $1,1,2,2, \cdots, 1986,1986$ be arranged in a row so that there is one number between the two 1s, two numbers between the two 2s, $\cdots$, and one thousand nine hundred and eighty-six numbers between the two 1986s? | ［Proof］ It is impossible to achieve. The conclusion is proven below using proof by contradiction.
By the problem statement, if it can be arranged, let the first number $k$ be written in the $a_{k}$ position, and the second number $k$ be written in the $b_{k}$ position, then it must be true that
$$b_{k}-a_{k}=k+1$$
Tak... | proof | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 734,755 |
$8 \cdot 15$ Let $a_{1}=3, b_{1}=100$, for $n \geqslant 1$
$$a_{n+1}=3^{a_{n}}, b_{n+1}=100^{b}{ }_{n}$$
Find the smallest positive integer $m$ such that $b_{m}>a_{100}$. | [Solution] Since $3^{3}a_{100}$, we have $m \leqslant 99$.
On the other hand, we can prove by induction that
$$a_{n+2}>6 b_{n}, n \geqslant 1$$
In fact, when $n=1$, we have $a_{3}=3^{27}>6 b_{1}=600$, so (1) holds. Suppose (1) holds for $n=k$. When $n=k+1$, we have
$$a_{k+3}=3^{a_{k+2}}>3^{6 b_{k}}>6 \cdot 100^{b_{k}}... | 99 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 734,757 |
8・16 Let $a_{1}, a_{2}, \cdots, a_{8}$ be 8 real numbers, not all zero, and let $C_{n}=a_{1}^{n}+a_{2}^{n}+\cdots+a_{8}^{n}, n=1,2,3, \cdots$ It is known that the sequence $\left\{C_{n}\right\}$ has infinitely many terms equal to 0. Find all natural numbers $n$ such that $C_{n}=0$. | [Solution] When $n$ is even, since $a_{1}, a_{2}, \cdots, a_{8}$ are not all 0, we have $C_{n}>0$. By the assumption, there exist infinitely many odd natural numbers $n_{1}0$ and when $\left|a_{i}\right|0$ and
$$0=\frac{C_{n_{k}}}{M_{1}^{n k}}=\sum_{i=1}^{8}\left(\frac{a_{i}}{M}\right)^{n_{k}}=\left(l_{1}-l_{1}^{\prime... | proof | Algebra | math-word-problem | Yes | Yes | inequalities | false | 734,758 |
8. 17 For every positive integer $x$, let $g(x)$ denote the largest odd divisor of $x$. Let
$$f(x)=\left\{\begin{array}{ll}
\frac{x}{2}+\frac{2}{g(x)}, & \text { if } x \text { is even, } \\
2^{\frac{x+1}{2}}, & \text { if } x \text { is odd. }
\end{array}\right.$$
Construct the sequence: $x_{1}=1, x_{n+1}=f\left(x_{n... | [Proof] For some positive integer $n$ and non-negative integer $k$, if $x_{n}=2^{k}$, then
$$x_{n+1}=3 \cdot 2^{k-1}, x_{n+2}=5 \cdot 2^{k-2}, \cdots, x_{n+k}=2^{k+1}$$
Thus, $x_{n+k+1}=2^{k+1}$. From this, it is easy to prove by induction that
$$x \frac{1}{2} k(k+1)+s=(2 s-1) 2^{k-s+1}$$
where $k=0,1,2,3, \cdots, s=... | 8253 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 734,759 |
8. 19 Given $S=\{1,2,3,4\}$. Let $a_{1}, a_{2}, \cdots, a_{k}$ be a sequence of numbers from $S$, and it includes all permutations of $(1,2,3,4)$ that do not end with 1, i.e., if $\left(b_{1}, b_{2}, b_{3}, b_{4}\right)$ is a permutation of $(1,2,3,4)$ and $b_{4} \neq 1$, then there exist $1 \leqslant i_{1}<i_{2}<i_{3}... | [Solution] Since $a_{1}, a_{2}, \cdots, a_{k}$ includes all permutations of $(1,2,3,4)$ where the second number is 1, there exist
$$1 \leqslant i_{1}<i_{2}<i_{3}<i_{4}<i_{5}<\cdots<i_{m} \leqslant k$$
such that $\left(a_{i_{1}}, a_{i_{2}}, a_{i_{3}}\right)$ is a permutation of $(2,3,4)$, $a_{i_{4}}=1$, and $a_{i_{5}},... | 11 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 734,761 |
8・20 Let $x_{1}, x_{2}, \cdots, x_{n}$ be a sequence of positive integers, each not greater than $M$, and satisfying
$$x_{k}=\left|x_{k-1}-x_{k-2}\right|, k=3,4, \cdots, n .$$
How many terms can such a sequence have at most? | [Solution] From (1), when $x_{k}=2$, we have $1, m, m-1,1, m-2, \cdots$, thus
$$\begin{array}{l}
l_{m}=3+l_{m-2} . \\
l_{m}=\left\{\begin{array}{ll}
\frac{3 m+1}{2}, & m \text { is odd, } \\
\frac{3 m+2}{2}, & m \text { is even. }
\end{array}\right.
\end{array}$$
For positive integers $m>1$, due to $m-1, m, 1, m-1, \c... | \sigma_{M}=\left\{\begin{array}{ll}
2, & M=1 ; \\
\frac{3 M+3}{2}, & M \text { is greater than } 1 \text { and is odd; } \\
\frac{3 M+2}{2}, & M \text { is even. }
\end{array}\right.} | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 734,762 |
8- 22 Given 3 numbers $x, y, z$. The absolute values of their pairwise differences are denoted as $\dot{x}_{1}, y_{1}, z_{1}$. The absolute values of the pairwise differences of $x_{1}, y_{1}, z_{1}$ are denoted as $x_{2}, y_{2}, z_{2}$. Continuing this process, after $n$ steps, it is found that $x_{n}=x, y_{n}=y, z_{n... | [Solution] Clearly, $x=x_{n} \geqslant 0, y=y_{n} \geqslant 0, \approx=z_{n} \geqslant 0$. For any two non-negative real numbers $a, b$, it is easy to see that
$$|a-b| \leqslant \max (a, b)$$
and the equality holds if and only if at least one of $a, b$ is 0. Therefore, at least one of $x, y, \approx$ is 0, and at leas... | (x, y)=(0,0) \text { or }(0,1) \text { or }(1,0) | Algebra | math-word-problem | Yes | Yes | inequalities | false | 734,763 |
8- 23 Given real numbers $\alpha, \beta$, such that $\alpha \beta>0$. Find the set of real numbers $r$ with the following property: there does not exist an infinite sequence $\left\{x_{n}\right\}$ satisfying
$$x_{0}=r, x_{n+1}=\frac{x_{n}+\alpha}{\beta x_{n}+1}, n=0,1,2, \cdots$$ | Let the required set be $M$, and the function
$$f(x)=\frac{x+\alpha}{\beta x+1} \quad\left(x \neq-\frac{1}{\beta}\right)$$
If $\alpha \beta=1$, then for any $x \neq-\frac{1}{\beta}$, we have
$$f(x)=\frac{x+\alpha}{\beta x+1}=\alpha=\frac{1}{\beta} \neq-\frac{1}{\beta},$$
Thus, for any $r \not \neq-\frac{1}{\beta}$, t... | M=\left\{r_{0}, r_{1}, r_{2}, \cdots\right\}, \text{ where } r_{n}=-\sqrt{\frac{\alpha}{\beta}} \frac{1+\lambda^{n+1}}{1-\lambda^{n+1}}, \lambda=\frac{1-\sqrt{\alpha \beta}}{1+\sqrt{\alpha \beta}} \text{ for } \alpha, \beta > | Algebra | math-word-problem | Yes | Yes | inequalities | false | 734,764 |
8・24 Let $r$ be a real number, and the sequence $\left\{x_{n}\right\}$ satisfies: $x_{0}=0, x_{1}=1$, and $x_{n+2}=$ $r x_{n+1}-x_{n}, n \geqslant 0$. For what value of $r$ does it hold that
$x_{1}+x_{3}+\cdots+x_{2 m-1}=x_{m}^{2}$, for any $m \geqslant 1$. | ［Solution］ The characteristic equation of the given recurrence relation is
$$\lambda^{2}-r \lambda+1=0,$$
Its roots are $\frac{1}{2}\left(r \pm \sqrt{r^{2}-4}\right)$.
If $r=2$, then the characteristic roots are the double root 1, in which case
$$x_{n}=n, n=0,1,2, \cdots$$
If $r=-2$, then $\lambda=-1$ is a double cha... | r=2 \text{ or } r=-2 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 734,765 |
8-25 Let $p_{0}>2$ be a given prime number, and the sequence $a_{0}, a_{1}, a_{2}, \cdots$ satisfies $a_{0}=0$, $a_{1}=1$, and
$$a_{n+1}=2 a_{n}+(a-1) a_{n-1}, n=1,2, \cdots$$
where the natural number $a$ is a parameter. Find the smallest value of $a$ that satisfies the following two conditions.
(i) If $p$ is a prime ... | ［Solution］ From the recurrence relation and $a_{0}=0, a_{1}=1$, we have
$$a_{n}=\frac{1}{2 \sqrt{a}}\left[(1+\sqrt{a})^{n}-(1-\sqrt{a})^{n}\right] .$$
Thus, for any odd prime $p$,
$$\begin{aligned}
a_{p} & =\frac{1}{2 \sqrt{a}}\left[\sum_{k=0}^{p} C_{p}^{k} a^{\frac{k}{2}}-\sum_{k=0}^{p}(-1)^{k} C_{p}^{k} a^{\frac{k}{... | not found | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 734,766 |
8-26 Given the sequence $\left\{a_{n}\right\}$ satisfies
$$a_{1}=\frac{1}{2}, a_{1}+a_{2}+\cdots+a_{n}=n^{2} a_{n}, n \geqslant 1 .$$
Find the general term formula for $a_{n}$. | [Solution] When $n \geqslant 2$, since
$$\begin{array}{l}
a_{1}+a_{2}+\cdots+a_{n}=n^{2} a_{n} \\
a_{1}+a_{2}+\cdots+a_{n-1}=(n-1)^{2} a_{n-1}
\end{array}$$
Therefore,
we can get
$$a_{n}=n^{2} a_{n}-(n-1)^{2} a_{n-1} .$$
Thus, we obtain $\quad a_{n}=\frac{1}{n(n+1)}, n=1,2,3, \cdots$ | a_{n}=\frac{1}{n(n+1)} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 734,767 |
8.27 Let the sequence of positive numbers $a_{0}, a_{1}, a_{2}, \cdots$ satisfy $a_{0}=a_{1}=1$ and
$$\sqrt{a_{n} a_{n-2}}-\sqrt{a_{n-1} a_{n-2}}=2 a_{n-1}, n=2,3, \cdots$$
Find the general term formula of the sequence. | [Solution] Since when $n \geqslant 2$,
$$\sqrt{a_{n}}=\sqrt{a_{n-1}}+\frac{2 a_{n-1}}{\sqrt{a_{n-2}}}$$
Let
$$\begin{array}{l}
b_{n}=\sqrt{\frac{a_{n}}{a_{n-1}}} \text {, we get } b_{1}=1, \text { and } \\
b_{n}=1+2 b_{n-1}, n=2,3, \cdots
\end{array}$$
It is easy to obtain
$$b_{n}=2^{n}-1, n=1,2, \cdots$$
Thus, we h... | a_{n}=\prod_{k=1}^{n}\left(2^{k}-1\right)^{2}, n=1,2,3, \cdots | Algebra | math-word-problem | Yes | Yes | inequalities | false | 734,768 |
Aug 28 Let the sequence $\left\{x_{n}\right\}$ satisfy: $x_{1}=2, x_{2}=3$, and
$$\left\{\begin{array}{l}
x_{2 m+1}=x_{2 m}+x_{2 m-1}, m \geqslant 1 \\
x_{2 m}=x_{2 m-1}+2 x_{2 m-2}, m \geqslant 2
\end{array}\right.$$
Find the general term formula for $x_{n}$. | [Solution] From the recurrence relation, when $m \geqslant 2$, we have
$$x_{2 m+1}=x_{2 m}+x_{2 m-1}=2 x_{2 m-1}+2 x_{2 m-2},$$
Also,
$$x_{2 m-2}=x_{2 m-1}-x_{2 m-3}$$
$$x_{2 m+1}=4 x_{2 m-1}-2 x_{2 m-3}$$
Therefore $\square$
Let $a_{m}=x_{2 m-1}$, then
$$a_{m+1}=4 a_{m}-2 a_{m-1}, m \geqslant 2$$
Similarly, let $b_... | null | Algebra | math-word-problem | Yes | Yes | inequalities | false | 734,769 |
8-29 Let the sequence $a_{1}, a_{2}, a_{3}, \cdots$ satisfy: $a_{1}=1$,
$$a_{n+1}=\frac{1}{16}\left(1+4 a_{n}+\sqrt{1+24 a_{n}}\right), n=1,2, \cdots$$
Find its general term formula. | [Solution] Let $b_{n}=\sqrt{1+24} a_{n}$, then for $n>1$, by the recursive relation of the sequence $\left\{a_{n}\right\}$, we have
$$\begin{aligned}
b_{n}^{2} & =1+24 a_{n}=1+\frac{24}{16}\left(1+4 a_{n-1}+\sqrt{1+24 a_{n-1}}\right) \\
& =\frac{1}{4}\left(10+24 a_{n-1}+6 \sqrt{1+24 a_{n-1}}\right) \\
& =\frac{1}{4}\le... | a_{n}=\frac{1}{3}+\left(\frac{1}{2}\right)^{n}+\frac{1}{3}\left(\frac{1}{2}\right)^{2 n-1} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 734,770 |
$$8 \cdot 30 \text { Let } \alpha=\frac{3-\sqrt{5}}{2}, f(n)=[\alpha n] \text {, and let }$$
$F(k)=\min \left\{n \in N ; f^{(k)}(n)>0\right\}$, for any $k \in N$, where $f^{(k)}=f \circ f \circ \cdots \circ f$ is the $k$-fold composition of $f$, find the expression for $F(k)$. | [Solution] Clearly, $f(1)=[\alpha]=0, f(2)=[3-\sqrt{5}]=0, f(3)=1$, so $F(1)=3$. Therefore, we have
$$f(f(n))>0 \Leftrightarrow f(n)=[\alpha n] \geqslant 3 .$$
Also, $[\alpha n] \geqslant 3 \Leftrightarrow n \alpha \geqslant 3 \Leftrightarrow n \geqslant \frac{3}{\alpha}=\frac{3(3+\sqrt{5})}{2} \Leftrightarrow n \geqs... | F(k)=\frac{1}{\sqrt{5}}\left(\frac{3+\sqrt{5}}{2}\right)^{k+1}-\frac{1}{\sqrt{5}}\left(\frac{3-\sqrt{5}}{2}\right)^{k+1} | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 734,771 |
8.31 $A$ and $E$ are a pair of opposite vertices of a regular octagon. A frog starts jumping from point $A$. If the frog is at any vertex other than $E$, it can jump to either of the two adjacent vertices. When it jumps to point $E$, it stops there. Let $e_{n}$ denote the number of different paths that reach point $E$ ... | [Solution] The eight vertices of a regular octagon are denoted as $A, B, C, D, E, F, G, H$ (as shown in the figure). The vertex $A$ is assigned the value $0$, $B, C, D$ are assigned the values $1, 2, 3$, and $H, G, F$ are assigned the values $-1, -2, -3$. A route from point $A$ to point $E$ in $k$ steps corresponds to ... | proof | Combinatorics | proof | Yes | Yes | inequalities | false | 734,772 |
8・32 Let $a_{1}=a_{2}=\frac{1}{3}$, and for $n=3,4,5, \cdots$
$$a_{n}=\frac{\left(1-2 a_{n-2}\right) a_{n-1}^{2}}{2 a_{n-1}^{2}-4 a_{n-2} a_{n-1}^{2}+a_{n-2}} .$$
(i) Find the general term formula for the sequence $\left\{a_{n}\right\}$.
(ii) Prove: $\frac{1}{a_{n}}-2$ is a perfect square of an integer. | ［Solution］ It is easy to prove by induction that
$$00$ . From the recursive formula, we can get
$$\frac{1}{a_{n}}=\frac{2 a_{n-1}^{2}-4 a_{n-2} a_{n-1}^{2}+a_{n-2}}{\left(1-2 a_{n-2}\right) a_{n-1}^{2}}=2+\frac{a_{n-2}}{\left(1-2 a_{n-2}\right) a_{n-1}^{2}}$$
That is, $b_{n}=\frac{1}{b_{n-2} a_{n-1}^{2}}=\frac{\left(b... | proof | Algebra | math-word-problem | Yes | Yes | inequalities | false | 734,773 |
8・33 The sequence of real numbers $a_{1}, a_{2}, \cdots, a_{n}, \cdots$ is defined by the following equation:
$$a_{n+1}=2^{n}-3 a_{n}, n=0,1,2, \cdots$$
(1) Find the expression for $a_{n}$ in terms of $a_{0}$ and $n$.
(2) Find $a_{0}$ such that for any positive integer $n, a_{n+1}>a_{n}$. | [Solution] (1) According to the definition of the sequence, we have
$$\begin{array}{l}
a_{n}=2^{n-1}-3 a_{n-1} \\
-3 a_{n-1}=-3 \cdot 2^{n-2}+3^{2} a_{n-2} \\
3^{2} a_{n-2}=3^{2} \cdot 2^{n-3}-3^{3} a_{n-3} \\
\cdots \cdots \cdots \cdots \\
(-1)^{n-1} 3^{n-1} a_{1}=(-1)^{n-1} 3^{n-1} 2^{0}+(-1)^{n} \cdot 3^{n} a_{0}
\e... | a_{0}=\frac{1}{5} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 734,774 |
8.34 The real number sequence $a_{0}, a_{1}, a_{2}, \cdots, a_{n}, \cdots$ satisfies the following equation: $a_{0}=a$, where $a$ is a real number,
$$a_{n}=\frac{a_{n-1} \sqrt{3}+1}{\sqrt{3}-a_{n-1}}, n \in N .$$
Find $a_{1994}$. | ［Solution］ After calculation, we can obtain
$$\begin{aligned}
a_{1} & =\frac{a \sqrt{3}+1}{\sqrt{3}-a} \\
a_{2} & =\frac{a_{1} \sqrt{3}+1}{\sqrt{3}-a_{1}}=\frac{(a \sqrt{3}+1) \sqrt{3}+(\sqrt{3}-a)}{\sqrt{3}(\sqrt{3}-a)-(a \sqrt{3}+1)} \\
& =\frac{a+\sqrt{3}}{1-a \sqrt{3}} \\
a_{3} & =\frac{a_{2} \sqrt{3}+1}{\sqrt{3}-a... | \frac{a+\sqrt{3}}{1-a \sqrt{3}} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 734,776 |
8-35 The sequence of integers $u_{0}, u_{1}, u_{2}, u_{3}, \cdots$ satisfies $u_{0}=1$, and for each positive integer $n$
$$u_{n+1} u_{n-1}=k u_{n}$$
where $k$ is some fixed positive integer. If
$$u_{2000}=2000$$
find all possible values of $k$. | ［Solution］ Let $u_{1}=u$. By the problem statement,
$$u_{2}=\frac{k u_{1}}{u_{0}}=k u$$
If $u=0$, then $u_{1}=0, u_{2}=0$. If $u_{t-1}=0$, then $k u_{t}=u_{t-1} \cdot$ $u_{t+1}=0$, hence $u_{t}=0$. Therefore, for any positive integer $n, u_{n}=0$. This contradicts $u_{2000}=2000$. Thus, $u \neq 0$.
By the problem sta... | k u=2000 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 734,777 |
8・36 Let the sequence $\left\{a_{n}\right\}$ have the sum of the first $n$ terms
$$S_{n}=2 a_{n}-1,(n=1,2, \cdots)$$
The sequence $\left\{b_{n}\right\}$ satisfies
$$b_{1}=3, b_{k+1}=a_{k}+b_{k},(k=1,2, \cdots)$$
Find the sum of the first $n$ terms of the sequence $\left\{b_{n}\right\}$. | [Solution] From $s_{n}=2 a_{n}-1$, we know,
$$\begin{array}{l}
a_{1}=2 a_{1}-1 \\
a_{1}=1
\end{array}$$
Also, $a_{k}=s_{k}-s_{k-1}=\left(2 a_{k}-1\right)-\left(2 a_{k-1}-1\right)=2 a_{k}-2 a_{k-1}$,
so $a_{k}=2 a_{k-1}$.
Thus, the sequence $\left\{a_{n}\right\}$ is a geometric sequence with the first term 1 and common... | 2^{n}+2 n-1 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 734,778 |
8.37 The sequence $a_{1}, a_{2}, a_{3}, \cdots$ satisfies
$a_{1}=a_{2}=1, a_{n}=\frac{a_{n-1}^{2}+2}{a_{n-2}}$, for all $n \geqslant 3$.
Prove that all numbers in the sequence are integers. | ［Proof］ It is easy to prove by induction that
$$a_{n}>0, n=1,2,3, \cdots$$
For $n \geqslant 3$, from the recurrence relation we have
$$\left\{\begin{array}{l}
a_{n} a_{n-2}=a_{n-1}^{2}+2 \\
a_{n+1} a_{n-1}=a_{n}^{2}+2
\end{array}\right.$$
Thus,
$$\begin{array}{l}
a_{n+1} a_{n-1}-a_{n} a_{n-2}=a_{n}^{2}-a_{n-1}^{2} \\... | proof | Algebra | proof | Yes | Yes | inequalities | false | 734,779 |
8.38 Prove: In the sequence
$$a_{k}=\left[2^{k} \sqrt{2}\right], k=1,2,3, \cdots$$
there are infinitely many composite numbers. | [Proof] From the assumption, we have
$$2^{k} \sqrt{2}=a_{k}+\lambda_{k}, k=1,2,3, \cdots$$
where $\lambda_{k}$ is an irrational number in $(0,1)$. If $0<\lambda_{k}<\frac{1}{2}$, then $a_{k+1}=2 a_{k}$ is even. If $\frac{1}{2}<\lambda_{k}<1$, then $a_{k+1}=2 a_{k}+1$ and
$$\lambda_{k+1}=2 \lambda_{k}-1=\lambda_{k}-\le... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 734,780 |
8-39 Let the non-zero sequence $a_{1}, a_{2}, \cdots$ satisfy: $a_{1}, a_{2}, \frac{a_{1}^{2}+a_{2}^{2}+b}{a_{1} a_{2}}$ are all integers, and
$$a_{n+2}=\frac{a_{n+1}^{2}+b}{a_{n}}, n=1,2, \cdots$$
where $b$ is a given integer. Prove that every term of the sequence $\left\{a_{n}\right\}$ is an integer. | [Proof] Let $n \geqslant 2$, from the recursive formula we get
thus
$$b=a_{n+2} a_{n}-a_{n+1}^{2}=a_{n+1} a_{n-1}-a_{n}^{2},$$
i.e., $\quad \frac{a_{n+2}+a_{n}}{a_{n+1}}=\frac{a_{n+1}+a_{n-1}}{a_{n}}, n=2,3, \cdots$
Let $c_{n}=\frac{a_{n+2}+a_{n}}{a_{n+1}}$, then
$$c_{n}=c_{1}=\frac{a_{3}+a_{1}}{a_{2}}=\frac{a_{1}^{2... | proof | Algebra | proof | Yes | Yes | inequalities | false | 734,781 |
$8 \cdot 40$ Prove that the sequence
$$a_{n}=\frac{(2+\sqrt{3})^{n}-(2-\sqrt{3})^{n}}{2 \sqrt{3}}, n=0,1,2, \cdots$$
is composed of integers, and find all $n$ such that $a_{n}$ is divisible by 3. | [Proof] Clearly, $a_{0}=0, a_{1}=1$. Since $2+\sqrt{3}$ and $2-\sqrt{3}$ are the roots of the equation $x^{2}-4 x +1=0$, the sequence $\left\{a_{n}\right\}$ satisfies the recurrence relation
$$a_{n+2}=4 a_{n+1}-a_{n}, n=0,1,2, \cdots$$
From this, it is easy to see that every term of $\left\{a_{n}\right\}$ is an intege... | 3\left|a_{n} \Leftrightarrow 3\right| n | Number Theory | proof | Yes | Yes | inequalities | false | 734,782 |
8・41 Let $a_{0}=0$, and for $n=0,1,2, \cdots$ we have
$$a_{n+1}=(k+1) a_{n}+k\left(a_{n}+1\right)+2 \sqrt{k(k+1) a_{n}\left(a_{n}+1\right)},$$
where $k$ is a given natural number. Prove that for $n \geqslant 0, a_{n}$ is an integer. | ［Proof］ By induction, it is easy to see that
$$0=a_{0}0$, so eliminating $a_{n+1}-a_{n-1}$ we get
$$a_{n+1}+a_{n-1}-(4 k+2) a_{n}-2 k=0$$
i.e., $a_{n+1}=(4 k+2) a_{n}-a_{n-1}+2 k, n=1,2, \cdots$
Furthermore, by $a_{0}=0, a_{1}=k, k$ being a natural number, it is immediately clear that for all non-negative integers $n,... | proof | Algebra | proof | Yes | Yes | inequalities | false | 734,783 |
8. 42 Given a sequence of positive numbers $\left\{a_{n}\right\}$ satisfying: $a_{1}=a_{2}=1, a_{3}=249$, and
$$a_{n+3}=\frac{1991+a_{n+2} a_{n+1}}{a_{n}}, n=1,2, \cdots$$
Prove that for all $n \in N, a_{n}$ is an integer. | [Proof] From the assumption, we have
$$\begin{array}{l}
a_{n} a_{n+3}=1991+a_{n+2} a_{n+1} \\
a_{n+1} a_{n+4}=1991+a_{n+3} a_{n+2}
\end{array}$$
Subtracting the two equations, we get
$$a_{n+1} a_{n+4}-a_{n} a_{n+3}=a_{n+3} a_{n+2}-a_{n+2} a_{n+1}$$
which simplifies to $\frac{a_{n+4}+a_{n+2}}{a_{n+3}}=\frac{a_{n+2}+a_... | proof | Algebra | proof | Yes | Yes | inequalities | false | 734,784 |
1. 110 If an integer is exactly the area of a right-angled triangle with integer side lengths (let $x, y, z \in N$ and $x^{2}+y^{2}=z^{2}$), it is called a Pythagorean number. Prove that for each positive integer $n$ greater than 12, there is a Pythagorean number between $n$ and $2n$. | [Proof] The positive integer solutions of the equation $x^{2}+y^{2}=z^{2}$ are
or $x=\left(a^{2}-b^{2}\right) c, y=2 a b c, z=\left(a^{2}+b^{2}\right) c$.
$$x=2 a b c, y=\left(a^{2}-b^{2}\right) c, z=\left(a^{2}+b^{2}\right) c \text {. }$$
Here $a, b, c$ are positive integers, and $a>b$. Thus, the area of the right t... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 734,786 |
8-44 Let the sequence $\left\{a_{n}\right\}$ satisfy:
$$a_{1}=1, a_{n+1}=\frac{a_{n}}{2}+\frac{1}{4 a_{n}}, n=1,2, \cdots$$
Prove: when $n>1$, $\sqrt{\frac{2}{2 a_{n}^{2}-1}}$ is a natural number. | ［Proof］By induction, it is easy to see that
$a_{n}>\frac{\sqrt{2}}{2}$, for any $n \geqslant 1$.
Let $b_{n}=\sqrt{\frac{2}{2 a_{n}^{2}-1}}$, since when $n \geqslant 2$,
$$2 a_{n}^{2}-1=2\left(\frac{a_{n-1}}{2}+\frac{1}{4 a_{n-1}}\right)^{2}-1=2\left(\frac{a_{n-1}}{2}-\frac{1}{4 a_{n-1}}\right)^{2} \text{,}$$
and by (1... | proof | Algebra | proof | Yes | Yes | inequalities | false | 734,787 |
8.45 Let $r$ be a positive integer, and the sequence $\left\{a_{n}\right\}$ satisfies:
$$a_{1}=1, a_{n+1}=\frac{n a_{n}+2(n+1)^{2 r}}{n+2}, n=1,2, \cdots$$
Prove that each $a_{n}$ is a positive integer, and find all $n$ such that $a_{n}$ is even. | [Proof] Since
$$(n+2) a_{n+1}=n a_{n}+2(n+1)^{2 r}$$
Therefore $\square$
$$(n+2)(n+1) a_{n+1}=(n+1) n a_{n}+2(n+1)^{2 r+1}$$
Let
$$\begin{array}{l}
b_{n}=(n+1) n a_{n}, n=1,2,3, \cdots \\
b_{n+1}=b_{n}+2(n+1)^{2 r+1}
\end{array}$$
Then $\square$
Thus, from $b_{1}=2$ we have
$$b_{n}=2 \sum_{k=1}^{n} k^{2 r+1}, n=1,2,... | proof | Algebra | proof | Yes | Yes | inequalities | false | 734,788 |
8-46 Let the sequence $a_{1}, a_{2}, a_{3}, \cdots$ satisfy
$$a_{1}=a_{2}=1, a_{n}=a_{n-1}+a_{n-2}, n=3,4,5, \cdots$$
Prove that for any natural number $k, a_{5 k}$ is divisible by 5. | ［Proof］ By induction. When $k=1$, we have
$$a_{5}=a_{4}+a_{3}=a_{3}+2 a_{2}+a_{1}=3 a_{2}+2 a_{1}=5$$
which shows that $5 \mid a_{5}$. Assume for some natural number $k$, $5 \mid a_{5 k}$.
Since
$$\begin{aligned}
a_{5 k+5} & =a_{5 k+4}+a_{5 k+3} \\
& =a_{5 k+3}+2 a_{5 k+2}+a_{5 k+1} \\
& =a_{5 k+2}+3 a_{5 k+1}+3 a_{5 ... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 734,789 |
8. 47 Prove that the sequence
$$a_{n}=\left(\frac{3+\sqrt{5}}{2}\right)^{n}+\left(\frac{3-\sqrt{5}}{2}\right)^{n}-2, n=1,2, \cdots$$
each term is a natural number, and when $n$ is even or odd, $a_{n}$ has the form $5 m^{2}$ or $m^{2}$, respectively, where $m \in N$. | [Proof] Clearly, $a_{1}=1, a_{2}=5$. Since $\frac{3+\sqrt{5}}{2}$ and $\frac{3-\sqrt{5}}{2}$ are the roots of the equation $x^{2}-3 x +1=0$, we have
$$a_{n+2}+2=3\left(a_{n+1}+2\right)-\left(a_{n}+2\right)$$
which means
$$a_{n+2}=3 a_{n+1}-a_{n}+2, n=1,2, \cdots$$
From this, it is easy to see that every term of the s... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 734,790 |
8.48 Given the sequence $\left\{a_{n}\right\}$ with the general term formula
$$a_{n}=[n \sqrt{2}], n=0,1,2, \cdots$$
Prove: $\left\{a_{n}\right\}$ has infinitely many terms that are perfect squares. | [Proof] For any positive odd number $m$, there exist natural numbers $x_{m}, y_{m}$, such that
$$(\sqrt{2}+1)^{m}=x_{m} \sqrt{2}+y_{m}$$
Since $(\sqrt{2}+1)^{m}+(\sqrt{2}-1)^{m}=2 x_{m} \sqrt{2}$, we have
$$(\sqrt{2}-1)^{m}=x_{m} \sqrt{2}-y_{m}$$
From this, we can derive $2 x_{m}^{2}-y_{m}^{2}=1$.
Thus, $y_{m}^{4}<2 ... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 734,791 |
8. 49 Let \( a_{n} \) be the number of natural numbers \( N \) such that the sum of the digits of \( N \) in its decimal representation is \( n \), and each digit can only be 1, 3, or 4. Prove that \( a_{2 n} \) is a perfect square. | [Proof] Clearly, $a_{1}=1, a_{2}=1, a_{3}=2, a_{4}=4$, and
$$a_{n}=a_{n-1}+a_{n-3}+a_{n-4}, n>4$$
It suffices to prove that for $n \geqslant 3$, $a_{2 n}$ is a perfect square. From (1), we have
$$\begin{aligned}
a_{2 n} & =a_{2 n-1}+a_{2 n-3}+a_{2 n-4} \\
& =a_{2 n-2}+a_{2 n-4}+a_{2 n-5}+a_{2 n-3}+a_{2 n-4}
\end{align... | proof | Combinatorics | proof | Yes | Yes | inequalities | false | 734,792 |
8-51 Let the sequence of integers $\left\{a_{n}\right\}$ satisfy:
$$a_{0}=0, a_{1}=1, a_{n}=2 a_{n-1}+a_{n-2}, n=2,3,4, \cdots$$
Prove: $2^{k}\left|a_{n} \Leftrightarrow 2^{k}\right| n, k=0,1,2, \cdots$ | [Proof] It is sufficient to prove the case for $k \geqslant 1$. From the recursive formula, we have
$$a_{n} \equiv a_{n-2}(\bmod 2), n=2,3,4, \cdots$$
Given $a_{0}=0, a_{1}=1$, we have
$$2\left|a_{n} \Leftrightarrow 2\right| n,$$
which means the conclusion holds for $k=1$. Next, we prove the case for $k>1$. From the ... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 734,794 |
8. 52 Define the function $f(n)$ on the set of positive integers as follows:
$$f(n)=\left\{\frac{n}{2}, n+3\right.$$
when $n$ is even,
when $n$ is odd,
(1) Prove that for any positive integer $m$, the sequence
$$a_{0}=m, a_{1}=f(m), \cdots, a_{n}=f\left(a_{n-1}\right), \cdots$$
always contains a term that is 1 or 3.
... | [Solution] (1) Since $\left\{a_{n}\right\}$ is a sequence of natural numbers, there must be a minimum value $a_{n_{0}}$. If $a_{n}>3$, when $a_{n}$ is even, $a_{n+1}=\frac{a_{n}}{2}<a_{n}$; when $a_{n}$ is odd, $a_{n+2}=\frac{a_{n}+3}{2}<a_{n}$. Therefore, $a_{n_{0}} \leqslant 3$. If $a_{n_{0}}=2$, then $a_{n_{0}+1}=1$... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 734,795 |
8. 53 Given $a_{1}=1, a_{2}=2$,
$a_{n+2}=\left\{\begin{array}{l}5 a_{n+1}-3 a_{n}, a_{n} \cdot a_{n+1} \\ a_{n+1}-a_{n}, a_{n} \cdot a_{n+1} \text { is even, }\end{array}\right.$ is odd. Prove that for all $n \in N, a_{n} \neq 0$. | [Proof] Since $a_{1}$ is odd, $a_{2}$ is even, and by the recursive formula, it is easy to see that in the sequence $\left\{a_{n}\right\}$, any two consecutive terms cannot both be even. It can be proven that $4 \nmid a_{n}, n=1,2,3, \cdots$, from which it follows that for any $n \in \mathbb{N}, a_{n} \neq 0$. In fact,... | proof | Algebra | proof | Yes | Yes | inequalities | false | 734,796 |
8.54 Given for any $n \in N, a_{n}>0$ and
$$\sum_{k=1}^{n} a_{k}^{3}=\left(\sum_{k=1}^{n} a_{k}\right)^{2}$$
Prove: $a_{n}=n, n=1,2,3, \cdots$ | [Proof] By induction. When $n=1$, from $a_{1}^{3}=a_{1}^{2}$ and $a_{1}>0$, we have $a_{1}=1$. Assume that for $n \leqslant m$, $a_{n}=n$. Since
$$\begin{array}{l}
\sum_{k=1}^{m} a_{k}^{3}+a_{m+1}^{3} \\
=\left(\sum_{k=1}^{m+1} a_{k}\right)^{2}=\left(\sum_{k=1}^{m} a_{k}\right)^{2}+2 a_{m+1}\left(\sum_{k=1}^{m} a_{k}\r... | proof | Algebra | proof | Yes | Yes | inequalities | false | 734,797 |
8. 55 A finite sequence $a_{1}, a_{2}, \cdots a_{n}$ is called $p$-equidistant if all sums of the following form are equal:
$$a_{k}+a_{k+p}+a_{k+2 p}+\cdots, k=1,2, \cdots, p$$
Prove that if a finite sequence with 50 terms is $p$-equidistant for $p=3,5,7,11,13,17$, then all terms of the sequence are 0. | [Proof] Let $p>1$ be a natural number, and the complex number $z$ satisfies
then $\square$
$$\begin{array}{c}
z^{p}=1 \text { and } z \neq 1, \\
1+z+z^{2}+\cdots+z^{p-1}=0 .
\end{array}$$
Thus, $\quad a_{1}+a_{1+p}+a_{1+2 p}+\cdots=a_{1}+a_{1+p} z^{p}+a_{1+2 p} z^{2 p}+\cdots$
$$\begin{array}{l}
z\left(a_{2}+a_{2+p}+... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 734,798 |
8-56 Let $n$ be a non-negative integer, and denote
$$(1+4 \sqrt[3]{2}-4 \sqrt[3]{4})^{n}=a_{n}+b_{n} \sqrt[3]{2}+c_{n} \sqrt[3]{4}$$
where $a_{n}, b_{n}, c_{n}$ are integers. Prove that if $c_{n}=0$, then $n=0$. | [Proof] Since
$$a_{n+1}+b_{n+1} \sqrt[3]{2}+c_{n+1} \sqrt[3]{4}=\left(a_{n}+b_{n} \sqrt[3]{2}+c_{n} \sqrt[3]{4}\right)(1+4 \sqrt[3]{2}-4 \sqrt[3]{4})$$
Therefore, $a_{n+1}=a_{n}-8 b_{n}+8 c_{n}, n=0,1,2, \cdots$
Given $a_{0}=1$, it follows that $a_{n}$ is always an odd number.
For every non-zero integer $k$, there exi... | proof | Algebra | proof | Yes | Yes | inequalities | false | 734,799 |
8. 57 Given the sequence $\left\{a_{n}\right\}$ with the general term
$$a_{n}=15 n+2+(15 n-32) \cdot 16^{n-1}, n=0,1,2, \cdots$$
(1) Prove that for each non-negative integer $n, 15^{3} \mid a_{n}$.
(2) Find all $n$ such that
$$1991\left|a_{n}, 1991\right| a_{n+1}, 1991 \mid a_{n+2} \text {. }$$ | [Solution] (1) Since
$$\begin{aligned}
2 a_{n+1}-a_{n}= & 2\left[15(n+1)+2+(15 n-17) \cdot 16^{n}\right]-[15 n+2+(15 n \\
& \left.-32) \cdot 16^{n-1}\right] \\
= & 15 n+30+2+(31 \times 15 n-32 \times 16) \cdot 16^{n-1} \\
= & a_{n+2}+(31 \times 15 n-32 \times 16) \cdot 16^{n-1}-(15 n-2) \cdot \\
& 16^{n+1} \\
= & a_{n+... | 89595k | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 734,800 |
8. 58 Prove: In any 3 infinite sequences of natural numbers
$$\begin{array}{l}
a_{1}, a_{2}, a_{3}, \cdots \\
b_{1}, b_{2}, b_{3}, \cdots \\
c_{1}, c_{2}, c_{3}, \cdots
\end{array}$$
there must exist different indices $p$ and $q$ such that
$$a_{p} \geqslant a_{q}, b_{p} \geqslant b_{q}, c_{p} \geqslant c_{q}$$ | [Proof] Since any finite set of natural numbers has a minimum, we can let $a_{n_{1}}=\min \left\{a_{n} \mid n \in N\right\}, a_{n_{2}}=\min \left\{a_{n} \mid n \in N, n>n_{1}\right\}, \cdots, a_{n_{k}}=$ $\min \left\{a_{n} \mid n \in N, n>n_{k-1}\right\}, \cdots$. Thus, $\left\{n_{k}\right\}$ is a subsequence of the na... | proof | Combinatorics | proof | Yes | Yes | inequalities | false | 734,801 |
8-59 Let $\left\{a_{n}\right\}$ be an infinite sequence of positive numbers, where $a_{1}$ can be arbitrarily chosen and for $n \geqslant 1$ we have
$$a_{n+1}^{2}=a_{n}+1$$
Prove: There exists $n$ such that $a_{n}$ is irrational. | [Proof] If $a_{1}>\frac{1+\sqrt{5}}{2}$, then
$$a_{1}^{2}-a_{1}-1>0$$
From this, we know
and $\quad a_{2}=\sqrt{a_{1}+1}>\sqrt{\frac{1+\sqrt{5}}{2}+1}=\frac{1+\sqrt{5}}{2}$.
$$a_{2}^{2}=a_{1}+1<a_{1}^{2},$$
Thus, by induction, $\left\{a_{n}\right\}$ is a strictly decreasing and bounded below infinite sequence of posi... | proof | Algebra | proof | Yes | Yes | inequalities | false | 734,802 |
8. 60 Given a two-ended infinite sequence $\cdots, a_{-n}, \cdots, a_{-1}, a_{0}, a_{1}, \cdots, a_{n}$, $\cdots$ satisfying
$$a_{k}=\frac{1}{4}\left(a_{k-1}+a_{k+1}\right) \text {, for any } k \in Z \text {. }$$
If there are two equal terms in this sequence, prove that there must be infinitely many pairs of equal ter... | ［Proof］Suppose there exists $k<p$ such that $a_{k}=a_{p}$. We can prove that
$$a_{k-1}=a_{p+1}$$
In fact, since $a_{k}=\frac{1}{4}\left(a_{k-1}+a_{k+1}\right)$, we have
$$a_{k+1}=4 a_{k}-a_{k-1}$$
By further recursion, we get
$$a_{k+2}=4 a_{k+1}-a_{k}=15 a_{k}-4 a_{k-1}$$
It is easy to prove by induction that
$$a_{k... | proof | Algebra | proof | Yes | Yes | inequalities | false | 734,803 |
$8 \cdot 61$ Prove: any natural number can be expressed as the sum of several different terms from the Fibonacci sequence $1,1,2,3$, $5,8,13, \cdots$
Express the above text in English, keeping the original text's line breaks and format, and output the translation result directly. | [Proof] The Fibonacci sequence is denoted as $a_{1}, a_{2}, a_{3}, \cdots$ then
$$a_{1}=a_{2}=1, a_{n}=a_{n-1}+a_{n-2}, n=3,4,5 \cdots$$
Using induction. When the natural number $m=1$, it is clearly $a_{1}=a_{2}=m$. Assume that for some $m \geqslant 1$, every natural number not exceeding $m$ can be expressed as the su... | null | Number Theory | proof | Yes | Yes | inequalities | false | 734,804 |
$8 \cdot 62$ Given the sequence $a_{1}, a_{2}, a_{3}, \cdots$ satisfies
$$a_{1}=a_{2}=1, a_{n}=a_{n-1}+a_{n-2}, n=3,4,5, \cdots .$$
Prove: In this sequence, the sum of any 8 consecutive terms is not equal to any term in the sequence. | [Proof] Clearly, $a_{1}, a_{2}, a_{3}, \cdots$ is a strictly increasing sequence of natural numbers. For any choice,
Since
$$\begin{aligned}
s & =a_{k}+a_{k+1}+\cdots+a_{k+7} \\
a_{k+8} & =a_{k+7}+a_{k+6}s
\end{aligned}$$
Therefore, $s$ does not equal any term in the sequence. | proof | Number Theory | proof | Yes | Yes | inequalities | false | 734,805 |
8・63 Let $x_{0}=0, x_{1}=1$, and $x_{n+1}=4 x_{n}-x_{n-1}, n=1,2,3, \cdots$ and $y_{0}=1, y_{1}=2$ and $y_{n+1}=4 y_{n}-y_{n-1}, n=1,2,3, \cdots$ Prove that for all integers $n \geqslant 0$,
$$y_{n}^{2}=3 x_{n}^{2}+1$$ | [Solution] From the recursive formula and the values of the first two terms, for any integer $n \geqslant 0$ we have
$$\begin{array}{l}
x_{n}=\frac{\sqrt{3}}{6}(2+\sqrt{3})^{n}-\frac{\sqrt{3}}{6}(2-\sqrt{3})^{n} \\
y_{n}=\frac{1}{2}(2+\sqrt{3})^{n}+\frac{1}{2}(2-\sqrt{3})^{n}
\end{array}$$
Thus,
$$\begin{array}{l}
x_{... | y_{n}^{2}=3 x_{n}^{2}+1 | Algebra | proof | Yes | Yes | inequalities | false | 734,806 |
1. 112 Suppose the decimal representation of the rational number $\frac{1}{s}$ is
$$\frac{r}{s}=0 . k_{1} k_{2} k_{3} \cdots$$
Prove that in the sequence
$$\begin{array}{l}
a_{1}=10 \frac{r}{s}-k_{1}, a_{2}=10^{2} \frac{r}{s}-\left(10 k_{1}+k_{2}\right), \\
a_{3}=10^{3} \frac{r}{s}-\left(10^{2} k_{1}+10 k_{2}+k_{3}\ri... | [Proof] Take the $m$-th digit of the infinite decimal representation of the rational number $\frac{r}{s}$, obtaining the number
$$0 . k_{1} k_{2} \cdots k_{m}=\frac{k_{1}}{10}+\frac{k_{2}}{10^{2}}+\cdots+\frac{k_{m}}{10^{m}}$$
This number is not greater than $\frac{r}{s}$, and adding $\frac{1}{10^{m}}$ to this number ... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 734,807 |
8・64 Let $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ be sequences of natural numbers, and satisfy
$$a_{n+1}=n a_{n}+1, b_{n+1}=n b_{n}-1, n \in N$$
Prove: There are at most a finite number of numbers that are terms in both $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$. | ［Proof］From the recursive relation, we can obtain
$$\begin{array}{c}
a_{2}=a_{1}+1 \\
\quad a_{3}=2 a_{2}+1=2\left(a_{1}+1\right)+1=2\left(a_{1}+1+\frac{1}{2}\right) \\
a_{4}=3 a_{3}+1=3 \cdot 2\left(a_{1}+1+\frac{1}{2}\right)+1=3!\left(a_{1}+1+\frac{1}{2!}+\frac{1}{3!}\right)
\end{array}$$
In general, it is easy to p... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 734,808 |
8.65 Starting from four given integers $a_{1}, b_{1}, c_{1}, d_{1}$, for all $n \in N$, define
$$\begin{array}{l}
a_{n+1}=\left|a_{n}-b_{n}\right|, \\
b_{n+1}=\left|b_{n}-c_{n}\right|, \\
c_{n+1}=\left|c_{n}-d_{n}\right|, \\
d_{n+1}=\left|d_{n}-a_{n}\right| .
\end{array}$$
Prove that there exists $k \in N$, such that
... | [Proof] Let $M_{n}=\max \left\{\left|a_{n}\right|,\left|b_{n}\right|,\left|c_{n}\right|,\left|d_{n}\right|\right\}$, then $M_{n}$ is a non-negative integer, and by the recursive relation, it is easy to see that
$$M_{n+1} \leqslant M_{n}, n=1,2, \cdots$$
From this, we know that there exists a natural number $k$ such th... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 734,809 |
\[
\begin{array}{c}
8 \cdot 66 \text { Let } a_{0}=1994, \text { and } \\
a_{n+1}=\frac{a_{n}^{2}}{a_{n}+1}, n=0,1,2, \cdots \\
\text { Prove: } 1994-n=\left[a_{n}\right], 0 \leqslant n \leqslant 998 .
\end{array}
\] | [Proof] It is clear that
$$0a_{n}-1>a_{n-1}-2>\cdots>a_{0}-(n+1)$$
Noting that $a_{0}=1994$, then for any non-negative integer $n$ we have
$$1994-n \leqslant a_{n}$$
On the other hand, when $0 \leqslant n \leqslant 997$, then by (1) we know
$$a_{n}>1994-997=997$$
Thus
$$a_{n+1}=a_{n}-1+\frac{1}{a_{n}+1}<a_{n}-1+\fra... | proof | Algebra | proof | Yes | Yes | inequalities | false | 734,810 |
8.67 Prove: From any sequence of real numbers $a_{1}, a_{2}, \cdots, a_{n}$, some numbers can be selected to satisfy the following three conditions:
(1) No three consecutive numbers are selected;
(2) From every three consecutive numbers, at least one is selected;
(3) The absolute value of the sum of all selected number... | [Proof] Without loss of generality, assume $a_{1}+a_{2}+\cdots+a_{n} \geqslant 0$, thus
$$\sum_{a_{k} \geqslant 0} a_{k} \geqslant \frac{1}{2} \sum_{i=1}^{n}\left|a_{i}\right|$$
For $i, k \in\{0,1,2\}$ and $i \neq k$, let
$$M_{i k}=\left\{a_{l} \mid l \equiv i(\bmod 3) \text { or } l \equiv k(\bmod 3) \text { and } a_... | proof | Combinatorics | proof | Yes | Yes | inequalities | false | 734,811 |
8-68 Let $a_{1}, a_{2}, a_{3}, \cdots$ be a sequence of real numbers such that for any $m$ and $n$,
$$\left|a_{m}+a_{n}-a_{m+n}\right| \leqslant \frac{1}{m+n}$$
Prove: $\left\{a_{k}\right\}$ is an arithmetic sequence. | [Proof] Taking $m=1$, then by the assumption we have
$$\left|\left(a_{n+1}-a_{n}\right)-a_{1}\right| \leqslant \frac{1}{n+1}, n=1,2,3, \cdots$$
Thus, $\lim _{n \rightarrow \infty}\left(a_{n+1}-a_{n}\right)=a_{1}$.
Taking $m=2$, from $\quad\left|\left(a_{n+2}-a_{n}\right)-a_{2}\right| \leqslant \frac{1}{n+2}, n=1,2,3, ... | proof | Algebra | proof | Yes | Yes | inequalities | false | 734,812 |
$8 \cdot 69$ Let $x_{1}, x_{2}, x_{3}, \cdots$ be a sequence of non-zero real numbers. Prove that the sequence is an arithmetic sequence if and only if for every $n \geqslant 2$,
$$\frac{1}{x_{1} x_{2}}+\frac{1}{x_{2} x_{3}}+\cdots+\frac{1}{x_{n-1} x_{n}}=\frac{n-1}{x_{1} x_{n}}$$
holds. | [Proof] Let $x_{1}, x_{2}, x_{3}, \cdots$ be an arithmetic sequence with a common difference $d$, and assume $d \neq 0$. For every $k \in \mathbb{N}$, we have
$$\frac{1}{x_{k} x_{k+1}}=\frac{1}{d}\left(\frac{1}{x_{k}}-\frac{1}{x_{k+1}}\right)$$
Therefore, for $n \geqslant 2$, we have
$$\sum_{k=1}^{n-1} \frac{1}{x_{k} ... | proof | Algebra | proof | Yes | Yes | inequalities | false | 734,813 |
8-70 When $n$ takes all natural numbers, prove that except for the terms of the sequence $a_{n}=3 n^{2}-2 n$,
$$f(n)=\left[n+\sqrt{\frac{n}{3}}+\frac{1}{2}\right]$$
takes all other natural numbers. | [Proof] Obviously, $f(n)=n+k_{n}, n=1,2,3, \cdots$ where $k_{n}=\left[\sqrt{\frac{n}{3}}+\frac{1}{2}\right]$. Since
$$2>\sqrt{\frac{n+1}{3}}+\frac{1}{2}-\sqrt{\frac{n}{3}}-\frac{1}{2}+1>k_{n+1}-k_{n} \geqslant 0$$
thus $\left\{k_{n}\right\}$ is a non-decreasing sequence and takes all natural numbers. For any natural n... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 734,814 |
$$\begin{array}{c}
8 \cdot 71 \quad \text { let } F_{1}=F_{2}=1, \\
F_{n+1}=F_{n}+F_{n-1}, n=2,3,4, \cdots \\
L_{1}=1, L_{2}=3, \text { and } \\
L_{n+1}=L_{n}+L_{n-1}, n=2,3,4, \cdots
\end{array}$$
Prove that for any natural numbers $n$ and $p$,
$$\left(\frac{L_{n}+\sqrt{5} F_{n}}{2}\right)^{p}=\frac{L_{n p}+\sqrt{5} ... | [Proof] The characteristic equation of the recurrence relation
$$\lambda^{2}-\lambda-1=0$$
has two distinct real roots $\alpha=\frac{1}{2}+\frac{\sqrt{5}}{2}, \beta=\frac{1}{2}-\frac{\sqrt{5}}{2}$. Therefore, according to the initial conditions of the two sequences, for $n=1,2,3, \cdots$
$$\begin{array}{l}
F_{n}=\frac... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 734,815 |
8.72 Given the integer sequence $\left\{a_{n}\right\}$ satisfies: $a_{1}=1, a_{2}=2$, and
$$a_{n+2}=\left\{\begin{array}{l}
5 a_{n+1}-3 a_{n}, a_{n} \cdot a_{n+1} \text { is even, } \\
a_{n+1}-a_{n}, a_{n} \cdot a_{n+1} \text { is odd, } \quad n=1,2,3, \cdots
\end{array}\right.$$
Prove:
(1) $\left\{a_{n}\right\}$ cont... | [Proof] From the recursive formula, we have
$$a_{n+2} \equiv a_{n+1}-a_{n}(\bmod 2), n=1,2,3, \cdots$$
From this, it is easy to see by induction that: when $n \equiv 0,1(\bmod 3)$, $a_{n}$ is odd, and when $n \equiv 2(\bmod 3)$, $a_{n}$ is even. Let $a_{3 n}=\alpha, a_{3 n+1}=\beta$, then $\alpha, \beta$ are both odd,... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 734,816 |
8・73 The sequence $\left\{x_{n} \mid n \in N\right\}$ is defined by the following formula:
$$x_{1}=2, n x_{n}=2(2 n-1) x_{n-1}, n=2,3, \cdots$$
Prove: For each positive integer $n, x_{n}$ is an integer. | [Proof] From the given conditions, we have
$$\begin{array}{ll}
2 x_{2}=6 x_{1}, & x_{2}=6=C_{4}^{2} \\
3 x_{3}=10 x_{2}, & x_{3}=20=C_{6}^{3}
\end{array}$$
Let \( x_{k}=C_{2 k}^{k} \)
Then for \( n=k+1 \), we have
$$(k+1) \cdot x_{k+1}=2(2 k+1) \cdot x_{2 k}$$
That is \(\square\)
$$\begin{aligned}
x_{k+1} & =\frac{2(... | proof | Algebra | proof | Yes | Yes | inequalities | false | 734,817 |
1. 113 If a number can be decomposed into the product of $k$ consecutive natural numbers greater than 1, then we say this number has property $p(k)$.
(1) Find a number $k$, for which there exists a number $N$ that simultaneously has properties $p(k)$ and $p(k+2)$.
(2) Prove that a number that simultaneously has propert... | [Solution] (1) $k=3$ is one of the numbers we are looking for. In fact, we have
$$720=2 \cdot 3 \cdot 4 \cdot 5 \cdot 6=8 \cdot 9 \cdot 10$$
(2) If a number $s$ has properties $p(2)$ and $p(4)$, then there exist natural numbers $m$ and $n$ such that
$$s=m(m+1)=n(n+1)(n+2)(n+3)$$
Thus, we have
$$s+1=m^{2}+m+1=\left(n^{... | proof | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 734,818 |
$8 \cdot 74$ Find the value of the smallest term in the following sequence:
$$a_{1}=1993^{1994^{1995}}, a_{n+1}=\left\{\begin{array}{ll}
\frac{1}{2} a_{n}, & \text { if } a_{n} \text { is even, } \\
a_{n}+7, & \text { if } a_{n} \text { is odd. }
\end{array}\right.$$ | [Solution] Clearly, all terms of the sequence are positive integers.
Since when $n$ is even,
$$a_{n+1}=\frac{1}{2} a_{n}<a_{n}$$
Therefore, the smallest term of the sequence must be odd.
Let the smallest term of the sequence be $a_{p}$, then $a_{p}$ is odd, and
$$\begin{array}{l}
a_{p+1}=a_{p}+7 \\
a_{p+2}=\frac{1}{2}... | 1 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 734,819 |
8. 75 Prove: The sequence $\left\{a_{n} \mid n \in N\right\}$ defined by $a_{n}=3^{n}-2^{n}$ does not contain three consecutive terms of a geometric progression. | [Proof] Since
$$\begin{aligned}
a_{n+1}-a_{n} & =\left(3^{n+1}-2^{n+1}\right)-\left(3^{n}-2^{n}\right) \\
& =2 \cdot 3^{n}-2^{n}>0
\end{aligned}$$
Therefore, for any positive integer \( n \),
$$a_{n+1}>a_{n}$$
When \( 1 \leqslant m < n \),
$$\begin{aligned}
a_{m} a_{2 n-m+1}-a_{n}^{2}= & \left(3^{m}-2^{m}\right)\left... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 734,820 |
8・76 Proof: All integers of the form $10017, 100117, 1001117, \cdots$ are divisible by 53. | [Solution] $10017=53 \times 189$, which shows that the first term is a multiple of 53.
Let the $n$-th term of this sequence be $a_{n}, n \in N$. Then
$$\begin{aligned}
a_{n+1}-a_{n} & =100 \underline{\underline{11 \cdots 17}}-100 \underline{11 \cdots 17}{ }_{n \uparrow 1}^{n+1 \uparrow 1} \\
& =901 \underline{00 \cdots... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 734,821 |
8. 77 The first 5 terms of a sequence are $1,2,3,4,5$, starting from the 6th term, each term is 1 less than the product of all previous terms. Prove: the product of the first 70 terms of this sequence is exactly equal to the sum of their squares. | [Proof] Let the $n$-th term of the sequence be $a_{n}$. Then,
$$\begin{array}{l}
a_{6}=119 \\
a_{n+1}=a_{1} a_{2} \cdots a_{n-1} a_{n}-1=a_{n}\left(a_{n}+1\right)-1, n \geqslant 6
\end{array}$$
$$a_{n+1}-a_{n}+1=a_{n}^{2}, n \geqslant 6$$
Thus, we have
$$\begin{aligned}
\sum_{n=1}^{70} a_{n}^{2} & =55+\sum_{n=6}^{70} ... | proof | Algebra | proof | Yes | Yes | inequalities | false | 734,822 |
8・ 78 The increasing sequence of integers that are divisible by 3 and are 1 less than a perfect square is $3,15,24,48, \cdots$ What is the remainder when the 1994th term of this sequence is divided by 1000? | [Solution] A number that is 1 less than a perfect square has the form $n^{2}-1=(n-1)$ $(n+1), n=2,3, \cdots$
If and only if $n$ is not divisible by 3, $n^{2}-1$ is a multiple of 3. Therefore, the $(2 k-1)$-th and $2 k$-th terms of this sequence are
$$(3 k-1)^{2}-1 \text { and }(3 k+1)^{2}-1 \text {. }$$
Thus, the 199... | 63 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 734,823 |
8・79 The sequence of positive numbers $\left\{a_{n}\right\}$ satisfies:
$$a_{1}=a_{2}=1, a_{3}=997, a_{n+3}=\frac{1993+a_{n+2} \cdot a_{n+1}}{a_{n}}, n \in N .$$
Prove that all $a_{n}$ are integers. | [Proof] Without loss of generality, let $a_{0}=2$. Then when $n=0$, we still have
$$a_{n+3}=\frac{1993+a_{n+2} \cdot a_{n+1}}{a_{n}} .$$
From the given, we get
$$a_{n+3} \cdot a_{n}=1993+a_{n+2} \cdot a_{n+1}$$
Replacing $n$ with $n+1$ gives
$$a_{n+4} \cdot a_{n+1}=1993+a_{n+3} \cdot a_{n+2}$$
Subtracting (2) from (1... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 734,824 |
8. 80 Given the sequence $\left\{a_{n}\right\}(n=1,2, \cdots)$ where $a_{1}=1, a_{2}=3, a_{3}=6$, and for $n>3$,
$$a_{n}=3 a_{n-1}-a_{n-2}-2 a_{n-3} .$$
Prove: For all natural numbers $n>3$, $a_{n}>3 \times 2^{n-2}$. | [Proof] From the recurrence relation, we get
$$a_{3}=6, a_{4}=13, a_{5}=27, a_{6}=56$$
Thus, we conjecture: when $n>3$, $a_{n}>2 a_{n-1}$.
We will prove this conjecture using mathematical induction.
It is clear that $a_{4}>2 a_{3}, a_{5}>2 a_{4}$.
Assume $a_{k}>2 a_{k-1}, a_{k-1}>2 a_{k-2}$. Then
$$\begin{aligned}
a_{... | proof | Algebra | proof | Yes | Yes | inequalities | false | 734,825 |
8. 81 Let $a_{1}, a_{2}, \cdots$ and $b_{1}, b_{2}, \cdots$ be two arithmetic sequences, and their sums of the first $n$ terms are $A_{n}$ and $B_{n}$, respectively. It is known that for all $n \in N$,
$$\frac{A_{n}}{B_{n}}=\frac{2 n-1}{3 n+1}$$
Try to write the expression for $\frac{a_{n}}{b_{n}}$ for all $n \in N$. | [Solution] Let the common difference of $a_{1}, a_{2}, \cdots$ be $d_{1}$; the common difference of $b_{1}, b_{2}, \cdots$ be $d_{2}$. Then
$$\begin{array}{l}
A_{n}=n a_{1}+\frac{n(n-1)}{2} d_{1} \\
B_{n}=n b_{1}+\frac{n(n-1)}{2} d_{2} \\
\frac{A_{n}}{B_{n}}=\frac{2 a_{1}+(n-1) d_{1}}{2 b_{1}+(n-1) d_{2}}
\end{array}$$... | \frac{4 n-3}{6 n-2} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 734,826 |
8. 82 Given a sequence of natural numbers $a_{1}, a_{2}, \cdots, a_{n}, \cdots$ where $a_{1}$ is not divisible by 5, and for each natural number $n$, there is the equation
$$a_{n+1}=a_{n}+b_{n}$$
where $b_{n}$ is the last digit of $a_{n}$. Prove: the sequence contains infinitely many terms that are powers of 2. | [Solution] From the given, we know that $b_{1} \notin\{0,5\}$. Therefore, $b_{2} \in\{2,4,6,8\}$, which leads to the conclusion that the sequence $b_{2}, b_{3}, \cdots$ has a period of 4, and has the following form:
$\cdots \cdots, 2,4,8,6,2,4,8,6, \cdots \cdots$ Thus, for any $n>1$, we have
$$a_{n+4}=a_{n}+(2+4+8+6)=a... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 734,827 |
$8 \cdot 83$ We use $(m, n)$ to denote the greatest common divisor of natural numbers $m$ and $n$. It is known that in the sequence of natural numbers $\left\{a_{i}\right\}$, for any $i \neq j$, we have $\left(a_{i}, a_{j}\right)=(i, j)$. Prove that for all $i \in N$, we have $a_{i}=i$.
Translate the above text into E... | [Proof] From the given,
$$\left(a_{i}, a_{2 i}\right)=(i, 2 i)=i$$
Therefore, for any natural number $i$, $a_{i}$ can be divided by $i$.
If for some natural number $i$, we have $a_{i}>i$, then from the given we get
$$\left(a_{a_{i}}, a_{i}\right)=\left(a_{i}, i\right)=i$$
From the previous proof, we can conclude that... | null | Number Theory | proof | Yes | Yes | inequalities | false | 734,828 |
1. The number of a bus ticket is a six-digit number. If the sum of the first three digits equals the sum of the last three digits, the ticket is called "lucky." Prove that the sum of all lucky ticket numbers is divisible by 13. | [Proof] If the lucky ticket number is $A$, then the ticket with the number $A^{\prime}=999999-A$ is also lucky. Moreover, $A^{\prime} \neq A$. Because
$$A+A^{\prime}=1001 \times 999=13 \times 77 \times 999$$
is divisible by 13, so the sum of all lucky ticket numbers is divisible by 13. | proof | Number Theory | proof | Yes | Yes | inequalities | false | 734,829 |
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