problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
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value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
1- 114 Proof: $m(m+1)$ is not a $k$-th power of any integer, where $m$ is a natural number, and $k$ is a natural number greater than 1. | [Proof] Suppose $m(m+1)$ is the $k$-th power of some natural number. Since $m$ and $m+1$ are coprime, each of them must be the $k$-th power of some number. Let $m=a^{k}, a \in N$. Then, we have
$$\begin{aligned}
(a+1)^{k} & >(a+1) a^{k-1} \\
& =a^{k}+a^{k-1} \\
& \geqslant m+1
\end{aligned}$$
The above inequality show... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 734,830 |
8-84 Prove that there exists an infinite bounded sequence $\left\{x_{n}\right\}$, such that for any different $m, k$ we have
$$\left|x_{m}-x_{k}\right| \geqslant \frac{1}{|m-k|}$$ | [Proof] For any natural number $m > k$, let
$$p = m - k, \quad q = [m \sqrt{2}] - [k \sqrt{2}].$$
Clearly, $p$ is a natural number, and $q$ is a non-negative integer, and
$$q \frac{1}{2 \sqrt{2} p + 1} > \frac{1}{4 p}.$$
Let $x_{n} = 4(n \sqrt{2} - [n \sqrt{2}]), \, n = 1, 2, 3, \cdots$. Clearly, $|x_{n}| < 4$. For an... | proof | Other | proof | Yes | Yes | inequalities | false | 734,831 |
8. 85 Prove: There exists a unique sequence of integers $a_{1}, a_{2}, a_{3}, \cdots$ such that
$$a_{1}=1, a_{2}>1, a_{n+1}^{3}+1=a_{n} a_{n+2}, n=1,2,3, \cdots$$ | [Proof] Uniqueness. Let the sequence $\left\{a_{n}\right\}$ satisfy the given conditions. For a natural number $k$, if $a_{k}>0, a_{k+1}>0$, then
$$a_{k+2}=\frac{a_{k+1}^{3}+1}{a_{k}}>0$$
By induction, it is easy to see that
$a_{n}>0$ for any $n \in N$.
Since $a_{3}=1+a_{2}^{3}$, we have
$$a_{4}=\frac{1+a_{3}^{3}}{a_{... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 734,832 |
8・86 Does there exist two real sequences $\left\{a_{n}\right\},\left\{b_{n}\right\}(n \in N)$, such that for every $n \in N$ and $x \in(0,1)$ we have
$$\frac{3}{2} \pi \leqslant a_{n} \leqslant b_{n}, \cos a_{n} x+\cos b_{n} x \geqslant-\frac{1}{n} ?$$ | [Solution] If there exist sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ that satisfy the given conditions, from $0<\frac{\pi}{a_{n}} \leqslant \frac{2}{3}<1$ we can get
$$\cos \frac{b_{n}}{a_{n}} \pi \geqslant 1-\frac{1}{n}$$
Furthermore, from $\frac{b_{n}}{a_{n}} \pi \geqslant \pi$ we know there exist $... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,833 |
8. 89 For a sequence of natural numbers with the following properties: the last term is 1969, and each term is greater than the square of the term preceding it, prove that the number of such sequences is less than 1969. | [Proof] Let $a_{1}, a_{2}, \cdots, a_{n}$ have the properties given in the problem. When $n=1$, there is only one case: $a_{1} = 1969$. Since
$$44^{2}<1969<45^{2}$$
when $n=2$, $a_{2}=1969, 1 \leqslant a_{1} \leqslant 44$, which means there are 44 cases. If $n=3$, then $a_{3}=1969, 2 \leqslant a_{2} \leqslant 44, 1 \l... | 741 | Number Theory | proof | Yes | Yes | inequalities | false | 734,835 |
$8 \cdot 90$ Let the sequence of non-zero real numbers $x_{1}, x_{2}, x_{3}, \cdots$ satisfy
$$x_{n}=\frac{x_{n-2} x_{n-1}}{2 x_{n-2}-x_{n-1}}, n=3,4,5, \cdots$$
Find the necessary and sufficient condition that $x_{1}$ and $x_{2}$ must satisfy so that the sequence has infinitely many integer terms. | [Solution] Since
$$\begin{aligned}
x_{n} & =\frac{x_{n-2} x_{n-1}}{2 x_{n-2}-x_{n-1}}=\frac{x_{n-2}}{2 x_{n-2}-\frac{x_{n-3} x_{n-2}}{2 x_{n-3}-x_{n-2}}} \cdot \frac{x_{n-3} x_{n-2}}{2 x_{n-3}-x_{n-2}} \\
& =\frac{x_{n-3} x_{n-2}}{3 x_{n-3}-2 x_{n-2}}
\end{aligned}$$
Thus, by induction, it is easy to prove that when $... | not found | Algebra | math-word-problem | Yes | Yes | inequalities | false | 734,836 |
8. 91 Given any natural number $a_{0}$, construct a sequence of natural numbers according to the following rules: $a_{0}$ is the first term. If $a_{n-1}$ has been constructed, the construction process ends when $a_{n-1} \leqslant 5$, resulting in a sequence with $n$ terms. If $a_{n-1} > 5$ but its last digit is no more... | [Solution] It is impossible. Let the constructed sequence $\left\{a_{n}\right\}$ be an infinite sequence of natural numbers. Then, when the last digit of $a_{n}$ is $\leqslant 5$,
$$10 a_{n+1}=a_{n}-a,$$
where $a$ is the last digit of $a_{n}$. This implies $a_{n+1}5$, and from $a_{n+1} = 9 a_{n} = 10 a_{n} - a_{n}$, w... | proof | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 734,837 |
8・92 The ternary sequence $\left(x_{n}, y_{n}, z_{n}\right), n=1,2,3, \cdots$ is constructed according to the following rules:
$$\begin{array}{l}
x_{1}=2, y_{1}=4, z_{1}=\frac{6}{7} \\
x_{n+1}=\frac{2 x_{n}}{x_{n}^{2}-1}, y_{n+1}=\frac{2 y_{n}}{y_{n}^{2}-1}, z_{n+1}=\frac{2 z_{n}}{z_{n}^{2}-1}, n \geqslant 1
\end{array... | [Solution] (1) Clearly, $x_{2}=\frac{4}{3}, y_{2}=\frac{8}{15}, z_{2}=-\frac{84}{13}$. If $\left(x_{k}, y_{k}, z_{k}\right), k=1,2, \cdots, n$ have all been constructed, where $n \geqslant 2$. Clearly, the constructed numbers are all rational, so they do not satisfy the quadratic equation $x^{2}-2 x-1=0$. Therefore, we... | proof | Algebra | proof | Yes | Yes | inequalities | false | 734,838 |
8.93 Find all arithmetic sequences $a_{1}, a_{2}, \cdots, a_{n}$, such that each term is an integer, and
$$\begin{array}{l}
a_{1}^{3}+a_{2}^{3}+a_{3}^{3}=a_{4}^{3} \\
a_{1}+a_{2}+\cdots+a_{n}=500
\end{array}$$ | [Solution] Let the common difference of the sequence be $d$. From the given conditions, we have
$$\begin{array}{l}
a_{1}^{3}+\left(a_{1}+d\right)^{3}+\left(a_{1}+2 d\right)^{3}=\left(a_{1}+3 d\right)^{3}, \\
n a_{1}+\frac{n(n-1)}{2} d=500,
\end{array}$$
where $a_{1}, d$ are integers, and $n$ is a natural number with $... | 60,80,100,120,140 \text{ and } 6,8,10,12, \cdots, 44 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 734,839 |
1. 115 Prove that for any natural number $n$, the sum $\sum_{k=0}^{n} 2^{3 k} C_{2 n+1}^{2 k+1}$ is not divisible by 5. | ［Proof］Let $x_{m}=\sum_{k} 2^{3 k} C_{m}^{2 k+1}(0<2 k+1 \leqslant m)$
$$y_{m}=\sum_{k} 2^{3 k} C_{m}^{2 k}(0 \leqslant 2 k \leqslant m)$$
Obviously, when $m$ is an odd number greater than 1, $x_{m}$ represents the sum in the problem.
According to the property of binomial coefficients $C_{r}^{s+1}+C_{r}^{s}=C_{r+1}^{s... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 734,840 |
8.94 Prove: Any four consecutive binomial coefficients $C_{n}^{r}, C_{n}^{r+1}, C_{n}^{r+2}, C_{n}^{r+3}$ cannot form an arithmetic sequence, where $n, r$ are positive integers, and $n \geqslant r+3$. | [Proof] If there exist $n, r (n \geqslant r+3)$, such that $C_{n}^{r}, C_{n}^{r+1}, C_{n}^{r+2}, C_{n}^{r+3}$ form an arithmetic sequence, then
$$\begin{array}{l}
2 C_{n}^{r+1}=C_{n}^{r}+C_{n}^{r+2} \\
2 C_{n}^{r+2}=C_{n}^{r+1}+C_{n}^{r+3}
\end{array}$$
That is, $2 \frac{n!}{(r+1)!(n-r-1)!}=\frac{n!}{r!(n-r)!}+\frac{n... | proof | Combinatorics | proof | Yes | Yes | inequalities | false | 734,841 |
8.96 Find all sequences of natural numbers $a_{1}, a_{2}, a_{3}, \cdots$ that satisfy the following two conditions:
(1) $a_{n} \leqslant n \sqrt{n}, n=1,2,3, \cdots$
(2) For any different $m, n$,
$$m-n \mid a_{m}-a_{n}$$ | ［Solution］ Let $\left\{a_{n}\right\}$ satisfy the conditions given in the problem, then
$$a_{1}=1, a_{2}=1 \text { or } 2 .$$
If $a_{2}=1$, by (2) we know that for any $n \geqslant 3$, there exist integers $x_{n}$ and $y_{n}$ such that
$$\left\{\begin{array}{l}
a_{n}=1+x_{n} \cdot(n-1) \\
a_{n}=1+y_{n} \cdot(n-2)
\end... | a_{n}=1 \text{, for any } n \geqslant 1 \text{; } a_{n}=n, \text{ for any } n \geqslant 1 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 734,842 |
8.97 Construct a sequence $x_{1}, x_{2}, x_{3}, \cdots$ from any real number $x_{1}$ satisfying
$$x_{n+1}=x_{n}\left(x_{n}+\frac{1}{n}\right), n=1,2, \cdots$$
Prove: There exists a unique real number $x_{1}$ such that the sequence constructed from it satisfies
$$0<x_{n}<x_{n+1}<1, n=1,2, \cdots$$ | [Proof] (1) Uniqueness. Suppose a real number $x_{1}$ satisfies the required conditions, then $00$ when $x>0$. For any fixed $x \geqslant 0$, the maximum root $y=f_{k}(x)$ of the equation $y\left(y+\frac{1}{k}\right)=x$ in terms of $y$, can be easily calculated to show that when $0\lambda_{k}, k=1,2,3, \cdots$
Since $... | proof | Algebra | proof | Yes | Yes | inequalities | false | 734,843 |
8.98 For what positive integer $n$, does there exist a permutation $x_{1}$, $x_{2}, \cdots, x_{n}$ of $1,2, \cdots, n$ such that $\left|x_{k}-k\right|, k=1,2, \cdots n$, are all different? | [Solution] Let $x_{1}, x_{2}, \cdots, x_{n}$ satisfy the conditions of the problem, then $|x_{1}-1|, |x_{2}-2|, \cdots, |x_{n}-n|$ must be a permutation of $0, 1, 2, \cdots, n-1$, so
$$\begin{aligned}
\sum_{k=1}^{n}\left|x_{k}-k\right| & =\sum_{k=0}^{n-1} k=\frac{n(n-1)}{2} \\
& \equiv \sum_{k=1}^{n} x_{k}-\sum_{k=1}^{... | proof | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 734,844 |
8.99 Given $n$ distinct natural numbers, try to prove for
(1) $n=5$, (2) $n=1989$
two cases that there exist infinitely many positive arithmetic sequences such that the first term is not greater than the common difference and the sequence contains exactly 3 or 4 of the given numbers. | [Proof] (1) Let the given 5 numbers be $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$. If their parity is not all the same, let there be $l$ odd numbers and $5-l$ even numbers, where $1 \leqslant l \leqslant 4$. In this case, there are only two scenarios. One is $3 \leqslant l \leqslant 4$, in which case the sequence of odd natur... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 734,845 |
8-100 (1) Does there exist a sequence of natural numbers $a_{1}, a_{2}, a_{3}, \cdots$ such that no term is equal to the sum of any other terms and satisfies
$$a_{n} \leqslant n^{10}, n=1,2,3, \cdots$$
(2) The problem is the same as (1) except that (1) is changed to
$$a_{n} \leqslant n \sqrt{n}, n=1,2,3, \cdots$$ | [Solution] (1) It exists. We construct a sequence $\left\{a_{n}\right\}$ that satisfies the required conditions in segments. First, let
$$b_{0}=1, \quad b_{n}=2^{3^{n-1}}, \quad n=1,2,3, \cdots$$
The first segment of $\left\{a_{n}\right\}$ consists of only one term, i.e., $a_{1}=1$. The second segment has $\frac{1}{2} ... | proof | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 734,846 |
8. 102 For $n=25$, construct a sequence of numbers $a_{1}, a_{2}, \cdots, a_{n}$ consisting of 0s and 1s, such that for any $0 \leqslant k \leqslant n-1$,
$$a_{1} a_{k+1}+a_{2} a_{k+2}+\cdots+a_{n-k} a_{n}$$
is odd;
(2) For some $n>1000$, prove: there exists a sequence $a_{1}, a_{2}, \cdots, a_{n}$ satisfying the cond... | [Solution] Let $A_{n}$ denote the sequence $a_{1}, a_{2}, \cdots, a_{n}$ composed of 0s and 1s. For $A_{n}$, define
$$P_{k}\left(A_{n}\right)=a_{1} a_{k+1}+a_{2} a_{k+2}+\cdots+a_{n-k} a_{n},$$
where $0 \leqslant k \leqslant n-1$. Since $a_{1}, a_{2}, \cdots, a_{n}$ take values 0 or 1, we have
$$\begin{aligned}
P\left... | proof | Combinatorics | proof | Yes | Yes | inequalities | false | 734,847 |
8. Does there exist a sequence of natural numbers in which each natural number appears exactly once, and for any $K=1,2,3, \cdots$ the sum of the first $K$ terms of the sequence is divisible by $K$? | [Solution] It exists.
In fact, we can construct a sequence $\left\{a_{n}\right\}$ that meets the conditions as follows:
$$\text { Let }\left\{\begin{array}{l}
a_{1}=1, \\
a_{n+1}=m\left[(n+2)^{t}-1\right]-S_{n}, \\
a_{n+2}=m .
\end{array}\right.$$
where $n=1,3,5, \cdots m$ is the smallest natural number different from... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 734,848 |
1. 116 The sum of squares of two consecutive integers can equal the square of another integer (such as $\left.3^{2}+4^{2}=5^{2}\right)$.
(1) Prove: When $m=3,4,5,6$, the sum of squares of $m$ consecutive integers cannot be the square of another integer.
(2) Find an example where the sum of squares of 11 consecutive int... | [Solution] (1) If $x$ is an integer, then $x^{2} \equiv 0$ or $1(\bmod 3), x^{2} \equiv 0$ or $1(\bmod 4)$. But
$$\begin{aligned}
& (x-1)^{2}+x^{2}+(x+1)^{2}=3 x^{2}+2 \equiv 2(\bmod 3) \\
& (x-1)^{2}+x^{2}+(x+1)^{2}+(x+2)^{2} \\
= & 4 x^{2}+4 x+6 \\
\equiv & 2(\bmod 4)
\end{aligned}$$
Therefore, the sum of the square... | 11(x^2 + 10) | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 734,849 |
8. 105 Proof: The arithmetic sequence with the first term 1 and common difference 729 must have infinitely many terms that are natural number powers of 10. | [Proof] Let this arithmetic sequence be $a_{1}, a_{2}, a_{3}, \cdots$
Then
$$a_{n+1}=1+729 n$$
Obviously, the necessary and sufficient condition for this sequence to have infinitely many terms that are natural powers of 10 is that there exist infinitely many natural numbers $n$ and $k$ such that
$$1+729 n=10^{k}$$
To... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 734,850 |
8. 106 Let $a, b$ be non-negative integers, and satisfy $a b \geqslant c^{2}$, where $c$ is an integer. Prove that there exist a number $n$ and integers $x_{1}, x_{2}, \cdots, x_{n} ; y_{1}, y_{2}, \cdots, y_{n}$, such that $\sum_{i=1}^{n} x_{i}^{2}=a, \sum_{i=1}^{n} y_{i}^{2}=b, \sum_{i=1}^{n} x_{i} y_{i}=c$. | [Proof] Let the above problem be denoted as $(a, b, c)$. Clearly, the proposition holds for $(a, b, c)$ if and only if it holds for $(a, b, -c)$. Therefore, we can assume $c \geqslant 0$. Furthermore, since the problem is symmetric with respect to $a$ and $b$, we can also assume $a \geqslant b$.
From $a b \geqslant c^... | proof | Algebra | proof | Yes | Yes | inequalities | false | 734,851 |
8-107 $a_{1}, a_{2}, \cdots, a_{m}$ are all non-zero numbers. Prove that if for any integer $k, k=0,1, \cdots, n(n<m-1)$, the following is satisfied:
$$a_{1}+a_{2} \cdot 2^{k}+a_{3} \cdot 3^{k}+\cdots+a_{m} \cdot m^{k}=0$$
then, in the sequence $a_{1}, a_{2}, \cdots, a_{m}$, there are at least $n+1$ pairs of adjacent ... | [Proof] Without loss of generality, let $a_{m}>0$. Otherwise, we can multiply all numbers in the sequence $a_{1}, a_{2}, \cdots, a_{m}$ by -1.
Take the sequence $b_{1}, b_{2}, \cdots, b_{n}$ such that
$$b_{i}=\sum_{j=0}^{n} c_{j} \cdot i^{j},$$
where $c_{0}, c_{1}, \cdots, c_{n}$ are arbitrary real numbers.
By the gi... | proof | Algebra | proof | Yes | Yes | inequalities | false | 734,852 |
8. 108 Question: Does there exist 12 geometric sequences such that the natural numbers $1,2,3, \cdots$, 100 are terms of these sequences? | ［Solution］ It does not exist.
In fact, if there are three different prime numbers $p_{1}, p_{2}, p_{3}$, where $p_{1}<p_{2}<p_{3}$, in the same geometric sequence, then we can set
$$p_{1}=a q^{k-1}, p_{2}=a q^{r-1}, p_{3}=a q^{m-1}$$
where $k, r, m$ are unequal natural numbers. Thus,
$$\begin{array}{l}
\frac{p_{2}}{p_... | proof | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 734,853 |
8-109 Does there exist an arithmetic sequence composed of increasing natural numbers with
(1) 11 terms;
(2) 10000 terms; $\square$
(3) infinitely many terms
such that the sum of the digits in decimal representation of each term also forms an increasing arithmetic sequence? | [Solution] (1) Exists. For example, $9,118,227,336,445,554,663,772,881$, 990,1099
(2) Exists. For example, first construct the following $10^{4} \times 10^{4}$ table:
\begin{tabular}{|c|c|c|c|c|c|}
\hline 00000 & 00001 & 00002 & $\ldots$ & 09998 & 09999 \\
\hline 00001 & 00002 & 00003 & $\ldots$ & 09999 & 10000 \\
\hli... | proof | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 734,854 |
8. 111 The first person writes down the numbers from 1 to 100 in some order as a sequence. The second person first writes down 50 different numbers from this sequence, and then asks the first person to write these 50 numbers in the order they appear in the original sequence; the second person then writes down another 5... | ［Solution］ At least 5 times are required.
In fact, to determine the order of the sequence $a_{1}, a_{2}, \cdots, a_{100}$, each adjacent pair $\left(a_{i}, a_{i+1}\right), i=1,2, \cdots, 99$, must appear at least once in the 50 numbers written each time; otherwise, the first person's answers to two different sequences
... | 5 | Combinatorics | proof | Yes | Yes | inequalities | false | 734,856 |
8. 113 Find a positive integer $k$, such that
(a) for any positive integer $n$, there does not exist $j$ satisfying $0 \leqslant j \leqslant n-k+1$, and $C_{n}^{j}$, $C_{n}^{i+1}, \cdots, C_{n}^{i+k-1}$ form an arithmetic sequence;
(b) there exists a positive integer $n$, such that there is $j$ satisfying $0 \leqslant ... | [Proof] Since any two numbers must form an arithmetic sequence, therefore $k \neq 1, k \neq 2$. Now consider three numbers: $C_{n}^{j-1}, C_{n}^{j}, C_{n}^{j+1}(1 \leqslant j \leqslant n-1)$
If they form an arithmetic sequence, then
$$2 C_{n}^{j}=C_{n}^{j-1}+C_{n}^{j+1}$$
This implies $n+2=(n-2 j)^{2}$
i.e., $n+2$ is ... | k=4, n=m^{2}-2(m \in N, m \geqslant 3) | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 734,858 |
$1 \cdot 117$ (1) Can the 10 numbers $0,1,2, \cdots, 9$ be placed on a circle so that the difference between any two adjacent numbers is 3, 4, or 5?
(2) Can the numbers $1,2, \cdots, 13$ be placed on a circle so that the difference between any two adjacent numbers is 3, 4, or 5? | [Solution] (1) Clearly, any two of the digits $0,1,2,8,9$ cannot be adjacent, meaning they should be placed on the circumference of the circle with at least one digit between them. However, among the written digits, only 2 can be adjacent to 7. Therefore, the digit 7 cannot be placed in any of the other 5 positions, so... | proof | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 734,859 |
8. 114 An infinite arithmetic sequence of positive integers contains one term that is a perfect square and another term that is a perfect cube. Prove that this sequence contains a term that is a perfect sixth power. | [Proof] Let the sequence be
$$\{a+i h \mid i=0,1,2, \cdots\}$$
which contains $x^{2}, y^{3}$ terms, where $x, y$ are integers.
We will use mathematical induction on $h$.
When $h=1$, the proposition is obviously true.
Assume that for any common difference less than $h$ and satisfying the given conditions, the propositi... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 734,860 |
8-115 Let the bounded integer sequence $a_{1}, a_{2}, a_{3}, \cdots$ satisfy
$$a_{n}=\frac{a_{n-1}+a_{n-2}+a_{n-3} a_{n-4}}{a_{n-1} a_{n-2}+a_{n-3}+a_{n-4}}, n \geqslant 5 .$$
Prove that there exists $l \in N$, such that $a_{l}, a_{l+1}, a_{l+2}, \cdots$ is a periodic sequence. | [Proof] From the assumption, there exists a positive integer $M$ such that
$$\begin{aligned}
& \left|a_{n}\right| \leqslant M, n=1,2,3 \cdots \\
\text { Let } & A=\left\{\left(a_{n}, a_{n+1}, a_{n+2}, a_{n+3}\right) ; n=1,2,3, \cdots\right\}
\end{aligned}$$ | proof | Algebra | proof | Yes | Yes | inequalities | false | 734,861 |
8・117 Let $\alpha_{n}$ and $\beta_{n}$ be the unit digits of $\left[(\sqrt{10})^{n}\right]$ and $\left[(\sqrt{2})^{n}\right]$, respectively, where $n=1,2,3, \cdots$ and $[x]$ denotes the greatest integer not exceeding $x$. Is $\left\{\alpha_{n}\right\}$ and $\left\{\beta_{n}\right\}$ a cyclic sequence, i.e., a periodic... | [Solution] $\left\{\alpha_{n}\right\}$ and $\left\{\beta_{n}\right\}$ are both non-repeating sequences. In fact, first consider $\left\{\alpha_{n}\right\}$. In decimal, let
$$\sqrt{10}=3 \cdot x_{1} x_{2} x_{3} \cdots$$
So $\alpha_{n}=\left\{\begin{array}{l}0, n=2 k, \\ x_{k}, n=2 k+1 .\end{array} \quad k=1,2,3, \cdot... | proof | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 734,862 |
8・118 Let the sequence $\left\{x_{n}\right\}$ satisfy
$$x_{1}=2, x_{n+1}=\left[\frac{3}{2} x_{n}\right], n=1,2,3 \cdots$$
Let $y_{n}=(-1)^{x_{n}}$, prove that $\left\{y_{n}\right\}$ is not a periodic sequence. | [Proof] From the assumption, we know that if $x_{n}$ is even, then $x_{n+1}=\frac{3}{2} x_{n}$, and if $x_{n}$ is odd, then $x_{n+1}=\frac{3}{2} x_{n}-\frac{1}{2}$. Therefore, for any natural number $n>m$, if $x_{n}$ and $x_{m}$ have the same parity, then
$$x_{n+1}-x_{m+1}=\frac{3}{2}\left(x_{n}-x_{m}\right)$$
If $\le... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 734,863 |
8-119 Given the sequence $a_{1}, a_{2}, a_{3}, \cdots$ satisfies $a_{2 n}=a_{n}$, for any $n \geqslant 1$, $a_{4 n+1}=1, a_{4 n+3}=0, \forall n \geqslant 0$.
Prove: This sequence is not a periodic sequence. | [Proof] By contradiction, let $a_{1}, a_{2}, a_{3}, \cdots$ be a periodic sequence with period $T=2^{r} q$, where $r$ is a non-negative integer and $q$ is a positive odd number. If there exists $n \geqslant 0$ such that $q=4 n+1$. Take $m \geqslant 0$ such that $m+n=4 k$, where $k$ is a non-negative integer. By assumpt... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 734,864 |
8・120 Given a sequence of positive integers in decimal notation $a_{1}, a_{2}, a_{3}, \cdots, a_{n}, \cdots$ that satisfies
$$a_{n}<a_{n+1} \leqslant 10 a_{n}, n=1,2,3 \cdots$$
Prove: The infinite decimal formed by writing down the numbers of this sequence one after another, $0 . a_{1} a_{2} a_{3} \cdots$, is not a re... | [Proof] By contradiction. Let $0 \cdot a_{1} a_{2} a_{3} \cdots$ be a repeating decimal with the shortest repeating cycle of length $T$. Since the sequence of natural numbers is strictly increasing, as $n \rightarrow \infty$, $a_{n} \rightarrow+\infty$. Also, $a_{n+1} \leqslant 10 a_{n}$, so the number of digits in $a_... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 734,865 |
8-122 Let $p(n)$ be the product of all digits in the decimal representation of the natural number $n$, and the sequence $\left\{n_{k}\right\}$ satisfies
$$n_{1} \in N, n_{k+1}=n_{k}+P\left(n_{k}\right), k=1,2,3 \cdots$$
Is $\left\{n_{k}\right\}$ unbounded? | The above not only proves that $\left\{n_{k}\right\}$ is bounded, but also proves that there exists $m$ such that
$$n_{m}=n_{m+k}, k=1,2,3, \cdots$$ | proof | Number Theory | proof | Yes | Yes | inequalities | false | 734,866 |
8. 123 For any non-negative integer $n$, let $S(n)=n-m^{2}$, where $m$ is the largest integer satisfying $m^{2} \leqslant n$. Let the sequence $\left\{a_{n}\right\}$ satisfy $a_{0}=A$, and
$$a_{n+1}=a_{n}+S\left(a_{n}\right), n \geqslant 0$$
For which non-negative integers $A$ does this sequence become constant from s... | [Solution] Clearly, when $A$ is a perfect square, the sequence is a constant sequence. Suppose $a_{n}$ is not a perfect square, then there exists $m$ such that
$$a_{n}=m^{2}+r$$
where $m, r$ are positive integers and $1 \leqslant r \leqslant 2 m$, i.e., $S\left(a_{n}\right)=r$. Thus,
$$a_{n+1}=m^{2}+2 r$$
Since $2 r ... | proof | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 734,867 |
8-124 Let the sequences of positive numbers $\left\{a_{k}\right\}$ and $\left\{b_{k}\right\}$ satisfy
(1) $a_{k}<b_{k}, k=1,2,3 \cdots$
(2) For any natural number $k$ and real number $x$,
$$\cos a_{k} x+\cos b_{k} x \geqslant-\frac{1}{k}$$
Prove: $\lim _{k \rightarrow \infty} \frac{a_{k}}{b_{k}}$ exists and find its v... | [Solution] Let $x=\frac{\pi}{b_{k}}$, then from (2) we have
$$\cos \frac{a_{k}}{b_{k}} \pi \geqslant 1-\frac{1}{k}, k=1,2,3 \cdots$$
By assumption (1), $0<\frac{a_{k}}{b_{k}} \pi<\pi$, thus
$$0<\frac{a_{k}}{b_{k}}<\frac{1}{\pi} \arccos \left(1-\frac{1}{k}\right) .$$
Since $\lim _{k \rightarrow \infty} \arccos \left(1... | 0 | Algebra | proof | Yes | Yes | inequalities | false | 734,868 |
8- 125 Given a positive number $a$ and a natural number $m$, let the sequence $\left\{x_{n}\right\}$ satisfy: for non-negative integers $n$,
$$x_{n+1}=\left(1-\frac{1}{m}\right) x_{n}+\frac{a}{m x_{n}^{m-1}},$$
where $x_{0}>0$ is arbitrarily given. Prove that: $\lim _{k \rightarrow \infty} x_{n}=\sqrt[m]{a}$. | [Proof] If $x_{n}>0$, then by the recursive relation and the mean value inequality, we have
$$\begin{aligned}
x_{n+1} & =\frac{m-1}{m} x_{n}+\frac{a}{m x_{n}^{m-1}} \\
& =\frac{1}{m}(\underbrace{x_{n}+\cdots+x_{n}}_{m-1 \uparrow}+\frac{a}{x_{n}^{m-1}}) \\
& \geqslant \sqrt[m]{a}
\end{aligned}$$
Thus, by induction, it ... | proof | Algebra | proof | Yes | Yes | inequalities | false | 734,869 |
8・126 Let the sequences $\left\{x_{n}\right\}$ and $\left\{y_{n}\right\}$ satisfy
$$y_{n}=x_{n-1}+2 x_{n}, n=2,3,4, \cdots$$
If $\left\{y_{n}\right\}$ converges, prove: $\left\{x_{n}\right\}$ also converges. | [Proof] Let $\alpha=\lim _{k \rightarrow \infty} y_{n}$, then for $n \geqslant 2$,
$$y_{n}-\alpha=x_{n-1}+2 x_{n}-\alpha=2\left(x_{n}-\frac{\alpha}{3}\right)+\left(x_{n-1}-\frac{\alpha}{3}\right)$$
Therefore, for $n=2,3,4, \cdots$ we have
$$\left|x_{n}-\frac{\alpha}{3}\right| \leqslant \frac{1}{2}\left|y_{n}-\alpha\ri... | proof | Algebra | proof | Yes | Yes | inequalities | false | 734,870 |
8. 127 Prove: When $n \rightarrow \infty$, the sequence $x_{n}=\sin n^{2}$ does not tend to 0. | [Proof] By contradiction. Suppose $\lim _{n \rightarrow \infty} \sin n^{2}=0$. From
$$|\sin (2 n+1)|=|\sin \left[(n+1)^{2}-n^{2}\right]|\leqslant| \sin (n+1)^{2}|+| \sin n^{2} |$$
we can see that $\quad \lim _{n \rightarrow \infty} \sin (2 n+1)=0$.
Furthermore, from $|\sin 2|=|\sin (2 n+3-2 n-1)|$
$\leqslant|\sin (2 n... | proof | Calculus | proof | Yes | Yes | inequalities | false | 734,871 |
8・128 Let $\left\{a_{n}\right\}$ be a sequence of positive numbers that is monotonically decreasing, and the sum of any finite number of its terms does not exceed 1. Prove that: $\lim _{n \rightarrow \infty} n a_{n}=0$.
| [Proof] Let $S_{0}=0, S_{n}=a_{1}+a_{2}+\cdots+a_{n}, n=1,2,3 \cdots$ Since $S_{n+1} - S_{n}=a_{n+1}>0, S_{n} \leqslant 1$, it follows that $\lim _{n \rightarrow \infty} S_{n}$ converges. For any natural number $n$, let
$$b_{n}=S_{n}-S\left[\frac{n}{2}\right]$$
Then, by the monotonicity of $\left\{a_{n}\right\}$, we h... | proof | Other | proof | Yes | Yes | inequalities | false | 734,872 |
8.129 Let the sequence $\left\{x_{n}\right\}$ satisfy $x_{1}=5$, and
$$x_{n+1}=x_{n}^{2}-2, n=1,2, \cdots$$
Find: $\lim _{n \rightarrow \infty} \frac{x_{n+1}}{x_{1} x_{2} \cdots x_{n}}$. | ［Solution］ It is easy to prove by induction
Since
$$\begin{aligned}
x_{n}>2, n & =1,2, \cdots \\
x_{n+1}^{2}-4 & =\left(x_{n}^{2}-2\right)^{2}-4=x_{n}^{2}\left(x_{n}^{2}-4\right) \\
& =x_{n}^{2} x_{n-1}^{2}\left(x_{n-1}^{2}-4\right) \\
& =\cdots=x_{n}^{2} x_{n-1}^{2} \cdots x_{1}^{2}\left(x_{1}^{2}-4\right)=21\left(x_... | \sqrt{21} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 734,873 |
8- 130 Let the sequence $\left\{a_{n}\right\}$ satisfy: $a_{1}=a_{2}=1$, and
$$a_{n+1}=a_{n}+\frac{a_{n-1}}{n(n+1)}, n=2,3, \cdots$$
Prove: this sequence converges. | [Proof] Clearly, for all $n \in N, a_{n}>0$, and $a_{n+1} \geqslant a_{n}$. Therefore, it suffices to prove that the sequence $\left\{a_{n}\right\}$ is bounded. From the recurrence relation, we have
$$\sum_{k=2}^{n} a_{k+1}=\sum_{k=2}^{n} a_{k}+\sum_{k=2}^{n} \frac{a_{k-1}}{k(k+1)}$$
Thus, $a_{n+1}=a_{2}+\sum_{k=2}^{n... | proof | Algebra | proof | Yes | Yes | inequalities | false | 734,874 |
8・131 Let the sequence $\left\{x_{n}\right\}$ satisfy: $x_{1}=1, x_{2}=-1$, and
$$x_{n+2}=x_{n+1}^{2}-\frac{1}{2} x_{n}, n \geqslant 1$$
Prove that $\lim _{n \rightarrow \infty} x_{n}$ converges and find this limit. | [Proof] It can be proven that if there exists $0<\delta<\frac{1}{2}$ and a natural number $k$ such that $\left|x_{k}\right| \leqslant \delta,\left|x_{k+1}\right| \leqslant \delta$, then $\lim _{n \rightarrow \infty} x_{n}=0$. In fact, from the recursive formula, we have
$\left|x_{k+2}\right| \leqslant\left|x_{k+1}\righ... | 0 | Algebra | proof | Yes | Yes | inequalities | false | 734,875 |
8. 132 For a decimal infinite fraction \( x \in [0,1] \), rearrange the first 5 digits after the decimal point to get a new decimal \( x_{1} \), then rearrange the 2nd to 6th digits after the decimal point of \( x_{1} \) to get \( x_{2} \). Continue this process, where \( x_{k+1} \) is obtained by rearranging the \( k+... | [Solution] (1) From the construction method of the sequence $\left\{x_{k}\right\}$, we know that for any natural number $k$, the first $k$ digits after the decimal point of the sequence $x_{k}, x_{k+1}, x_{k+2}, \cdots$ remain unchanged, i.e., if we denote
$$x_{k}=0, x_{k 1} x_{k 2} x_{k 3} \cdots, k=1,2,3, \cdots$$
t... | proof | Other | proof | Yes | Yes | inequalities | false | 734,876 |
8・133 Let $x_{n}=(1+\sqrt{2}+\sqrt{3})^{n}, n=1,2,3, \cdots$ If $x_{n}$ is expressed as $x_{n}=q_{n}+r_{n} \sqrt{2}+s_{n} \sqrt{3}+t_{n} \sqrt{6}$, where $q_{n}, r_{n}, s_{n}$, and $t_{n}$ are all integers, find the following limits:
$$\lim _{n \rightarrow \infty} \frac{r_{n}}{q_{n}}, \lim _{n \rightarrow \infty} \frac... | [Solution] Let $u_{1}=1+\sqrt{2}+\sqrt{3}, u_{2}=1-\sqrt{2}+\sqrt{3}, u_{3}=1+\sqrt{2}-\sqrt{3}$, $u_{4}=1-\sqrt{2}-\sqrt{3}$, it is easy to see that
$$\begin{array}{l}
u_{1}^{n}=q_{n}+r_{n} \sqrt{2}+s_{n} \sqrt{3}+t_{n} \sqrt{6}, \\
u_{2}^{n}=q_{n}-r_{n} \sqrt{2}+s_{n} \sqrt{3}-t_{n} \sqrt{6}, \\
u_{3}^{n}=q_{n}+r_{n}... | \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{6}} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 734,877 |
$1 \cdot 119$ (1) Prove: The numbers $1,2, \cdots, 32$ can be arranged in some order such that the half of the sum of any two numbers is not equal to any number between these two numbers.
(2) Can the numbers $1,2,3, \cdots, 100$ be arranged in some order such that the half of the sum of any two numbers is not equal to ... | [Solution] (1) First, we write all the numbers in a row, placing even numbers $2k_{1}$ on the left half and odd numbers $2k_{1}+1$ on the right half. The sum of any two numbers from different halves, divided by 2, is not an integer, and thus does not belong to these two halves.
Next, we divide each half into two parts... | proof | Combinatorics | proof | Yes | Yes | inequalities | false | 734,878 |
8-134 Let the infinite sequence $\left\{a_{n}\right\}$ satisfy $\lim _{n \rightarrow \infty}\left(a_{n+1}-\frac{1}{2} a_{n}\right)=0$. Prove: $\lim _{n \rightarrow \infty} a_{n}=0$. | [Solution] Since $\lim _{n \rightarrow \infty}\left(a_{n+1}-\frac{1}{2} a_{n}\right)=0$, according to the definition of limits, for any $\varepsilon > 0$, there exists a positive integer $N$ such that when $n \geqslant N$, we have $\left|a_{n+1}-\frac{1}{2} a_{n}\right|<\frac{\varepsilon}{4}$. For a fixed $N$, there ex... | proof | Calculus | proof | Yes | Yes | inequalities | false | 734,879 |
$$\begin{array}{l}
\text { 8・135 Let } a_{0}>0 \text { and } \\
\\
\qquad a_{n+1}=\frac{a_{n}^{2}-1}{n+1}, n \geqslant 0 .
\end{array}$$
Prove that there exists a real number $a>0$ such that
(i) if $a_{0} \geqslant a$, then as $n \rightarrow+\infty$, $a_{n} \rightarrow+\infty$.
(ii) if $a_{0}<a$, then as $n \rightarro... | [Proof] If $a_{n} \geqslant n+2$, then
$$a_{n+1} \geqslant \frac{(n+2)^{2}-1}{n+1}=\frac{n^{2}+4 n+3}{n+1}=n+3 \text {. }$$
Thus, if $a_{0} \geqslant 2$, by induction we can prove that
$$a_{n} \geqslant n+2 \text {, for all } n \geqslant 0 \text {. }$$
From this, we can conclude that as $n \rightarrow+\infty$, $a_{n}... | a=2 | Algebra | proof | Yes | Yes | inequalities | false | 734,880 |
8-137 Let the sequence of positive numbers $\left\{a_{n}\right\}$ satisfy
$$a_{n+1} \leqslant \sqrt[3]{a_{n} a_{n-1}^{2}}, n=2,3, \cdots$$
Prove: The sequence converges. | [Proof] Since $a_{n+1} \leqslant \sqrt[3]{a_{n} a_{n-1}^{2}}$, we have
$$a_{n+1} \leqslant \max \left\{a_{n-1}, a_{n}\right\}$$
For $n \geqslant 2$, let $b_{n}=\max \left\{a_{n-1}, a_{n}\right\}$. Then from (1) we get
$$b_{n+1}=\max \left\{a_{n}, a_{n+1}\right\} \leqslant \max \left\{a_{n}, a_{n-1}\right\}=b_{n}$$
Th... | proof | Algebra | proof | Yes | Yes | inequalities | false | 734,881 |
8. 138 Let $a(n)$ denote the number of 1's in the binary representation of the positive integer $n$. Prove:
(1) $a\left(n^{2}\right) \leqslant \frac{1}{2} a(n)(a(n)+1)$.
(2) The equality in the above inequality can hold for infinitely many positive integers.
(3) There exists a sequence $\left\{n_{k}\right\}$ such that
... | [Proof] Clearly, $a(1)=1$, and for any non-negative integer $k$ and positive integer $n$, we have
$$a\left(2^{k} n\right)=a(n) .$$
Furthermore, it is easy to see that when $n$ is even,
$$a(n+1)=a(n)+1,$$
and when $n$ is odd,
$$a(n+1) \leqslant a(n) .$$
Similarly, we can obtain
$$a\left(n+2^{k}\right) \leqslant a(n)+... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 734,882 |
8・ $140 a$ is a real number in $(0,1)$, consider the sequence $\left\{x_{n} \mid n=0,1,2, \cdots\right\}$, where
$$x_{0}=a, x_{n}=\frac{4}{\pi^{2}}\left(\arccos x_{n-1}+\frac{\pi}{2}\right) \cdot \arcsin x_{n-1}, n=1,2,3, \cdots \text { find }$$
Prove: As $n$ approaches infinity, the sequence $\left\{x_{n} \mid=0,1,2 ... | [Proof] Note
$\arccos x_{n-1}=\theta, \arcsin x_{n-1}=\varphi$,
where $\theta \in[0, \pi], \varphi \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. Therefore, we have
$$\cos \theta=x_{n-1}=\sin \varphi=\cos \left(\frac{\pi}{2}-\varphi\right)$$
Noting that $\frac{\pi}{2}-\varphi \in[0, \pi], \theta \in[0, \pi]$, so
$$\t... | 1 | Algebra | proof | Yes | Yes | inequalities | false | 734,884 |
8-141 The sequence $\left\{a_{n} \mid n \in N\right\}$ is defined by the following formula: $a_{1}=1994$, $a_{n+1}=\frac{a_{n}^{2}}{2\left[a_{n}\right]+21}, n \in N$.
Here $\left[a_{n}\right]$ is the greatest integer not exceeding $a_{n}$.
(1) Prove that $a_{12}<1$.
(2) Prove that this sequence is convergent, and find ... | [Solution] (1) Clearly, $a_{n}>0(n \in \mathbb{N})$. Using $\left[a_{n}\right]>a_{n-1}$, we get
$$\begin{aligned}
a_{n+1} & =\frac{a_{n}^{2}}{2\left[a_{n}\right]+21}>0$, thus it is sufficient that
$$00 \\
s+\sqrt{s^{2}+19 s}>2 s+5
\end{array}$$
Therefore, when $s \geqslant 3$, if $a_{n}>s$, then $a_{n+1}>s$.
From $a_{... | 10 | Algebra | proof | Yes | Yes | inequalities | false | 734,885 |
8-142 (1) Prove: The equation $x^{3}+x^{2}+x=a$ has only one real solution for each real number $a$.
(2) The sequence of real numbers $\left\{x_{n} \mid n \in N\right\}$ satisfies
$$x_{n+1}^{3}+x_{n+1}^{2}+x_{n+1}=x_{n} \text {, and } x_{1}>0 \text {. }$$
Prove: This sequence converges, and find its limit. | [Proof] (1) Let
$$f(x)=x^{3}+x^{2}+x-a$$
For $x>y$, we have
$$\begin{aligned}
f(x)-f(y) & =\left(x^{3}+x^{2}+x-a\right)-\left(y^{3}+y^{2}+y-a\right) \\
& =\left(x^{3}-y^{3}\right)+\left(x^{2}-y^{2}\right)+(x-y) \\
& =(x-y)\left(x^{2}+x y+y^{2}+x+y+1\right) \\
& =(x-y)\left[(x+y)^{2}+(x+y)+1-x y\right]
\end{aligned}$$
... | 0 | Algebra | proof | Yes | Yes | inequalities | false | 734,886 |
$8 \cdot 143$ The infinite sequence $x_{n}$ is defined by the following rule:
$$x_{n+1}=|1-| 1-2 x_{n}|| \text {, and } 0 \leqslant x_{1} \leqslant 1 \text {. }$$
(1) Prove: The sequence becomes periodic from some term onwards if and only if $x_{1}$ is a rational number.
(2) How many different values of $x_{1}$ exist s... | ［Solution］(1) From the given, we have
$$x_{n+1}=\left\{\begin{array}{l}
2 x_{n}, \text { if } 0 \leqslant x_{n}<\frac{1}{2} \\
2-2 x_{n}, \text { if } \frac{1}{2} \leqslant x_{n} \leqslant 1
\end{array}\right.$$
If $x_{1}$ is a rational number, we can set $x_{1}=\frac{p}{q},(p, q)=1, p \in N, q \in N$. Note that for a... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 734,887 |
1. 120 Given an infinite arithmetic sequence, each term of which is a positive integer, and one of its terms is a perfect square, prove: this sequence has infinitely many perfect squares. | [Proof] Let the common difference of the series be $d$, and let one of its terms be $a=m^{2}$, where $m$ is a natural number.
Since when $k$ is any natural number, the number
$$\begin{aligned}
(m+k d)^{2} & =m^{2}+2 m k d+k^{2} d^{2} \\
& =a+d\left(2 k m+k^{2} d\right)
\end{aligned}$$
are terms of the series, therefo... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 734,888 |
$8 \cdot 146$ Let $x_{n}=\sqrt{2+\sqrt[3]{3+\cdots+\sqrt[n]{n}}}, n=2,3, \cdots$ Prove that: $x_{n+1}-x_{n}<\frac{1}{n!}, n=2,3, \cdots$ | [Proof] For $2 \leqslant k \leqslant n$, let
$$\begin{array}{l}
b_{k}=\sqrt[k]{k+\sqrt[k+1]{k+1+\cdots+\sqrt[n]{n}}}, \\
c_{k}=a_{k}^{k-1}+a_{k}^{k-2} b_{k}+\cdots+a_{k} b_{k}^{k-2}+b_{k}^{k-1} .
\end{array}$$
Obviously, $x_{n+1}=a_{2}, x_{n}=b_{2}$, and
$$\begin{aligned}
x_{n+1}-x_{n} & =a_{2}-b_{2}=\frac{a_{3}-b_{3}... | proof | Algebra | proof | Yes | Yes | inequalities | false | 734,891 |
8・147 Let the sequence of positive real numbers $\left\{a_{k}\right\}$ satisfy
$$a_{1} \geqslant 1, a_{k+1}-a_{k} \geqslant 1, k=1,2,3, \cdots$$
Prove: For all natural numbers $n$,
$$\sum_{k=1}^{n} a_{k}^{-\frac{1}{1987} a_{k+1}^{-1}}<1987$$ | [Proof] Since $a_{k} \geqslant k, k=1,2,3, \cdots$
it suffices to prove that for all natural numbers $n$,
$$\sum_{k=1}^{n} k^{-\frac{1}{1987}}(k+1)^{-1}<1987$$
For any natural number $k$, let $\alpha=(k+1)^{\frac{1}{1987}}, \beta=k^{\frac{1}{1987}}$. Then
$$\begin{aligned}
1 & =(k+1)-k=\alpha^{1987}-\beta^{1987} \\
& ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,892 |
8・148 Let $x_{0}=10^{9}, x_{n}=\frac{x_{n-1}^{2}+2}{2 x_{n-1}}, n=1,2,3, \cdots$ Prove: $0<x_{36}-\sqrt{2}<10^{-9}$. | ［Proof］It is easy to prove by induction that
$$x_{n}>\sqrt{2}, n=0,1,2, \cdots$$
From the recursive formula, we have
$$\begin{aligned}
0<x_{n}-\sqrt{2} & =\frac{x_{n-1}^{2}+2}{2 x_{n-1}}-\sqrt{2}=\frac{1}{2}\left(\sqrt{x_{n-1}}-\sqrt{\frac{2}{x_{n-1}}}\right)^{2} \\
& =\frac{1}{2 x_{n-1}}\left(x_{n-1}-\sqrt{2}\right)^... | proof | Algebra | proof | Yes | Yes | inequalities | false | 734,893 |
8-149 Given a strictly increasing unbounded sequence of positive numbers $a_{1}, a_{2}, \cdots$, prove:
(1) There exists a natural number $k_{0}$ such that for all $k \geqslant k_{0}$,
$$\frac{a_{1}}{a_{2}}+\frac{a_{2}}{a_{3}}+\cdots+\frac{a_{k}}{a_{k+1}}<k-1 .$$
(2) When $k$ is sufficiently large,
$$\frac{a_{1}}{a_{2}... | [Proof] Let $s_{k}=\frac{a_{1}}{a_{2}}+\frac{a_{2}}{a_{3}}+\cdots+\frac{a_{k}}{a_{k+1}}$. It can be proven that for any $M>0$, there exists a natural number $k_{0}$ such that when $k \geqslant k_{0}$, we have
$$S_{k}<k-M$$
This immediately implies that (1) and (2) both hold.
Indeed, since $a_{1}, a_{2}, \cdots$ is inc... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,894 |
8. 152 The students of the school mathematics group made a calculator, which can transform a quadruple $(a, b, c, d)$ into $(a-b, b-c, c-d, d-a)$ when a button is pressed. Prove that if the numbers in the initial quadruple are not all equal, then after pressing the button a finite number of times, the numbers in the re... | [Proof] Let the initial quadruple be $(a, b, c, d)$, and after pressing the button $n$ times, the resulting quadruple is $\left(a_{n}, b_{n}, c_{n}, d_{n}\right)$. Clearly,
$$a_{n}+b_{n}+c_{n}+d_{n}=0 \text {, for any } n \geqslant 1 \text {. }$$
Let $s_{n}=a_{n}^{2}+b_{n}^{2}+c_{n}^{2}+d_{n}^{2}$, and by the fact tha... | proof | Algebra | proof | Yes | Yes | inequalities | false | 734,896 |
8. 153 Let $x_{1}, x_{2}, x_{3}, \cdots$ be a decreasing sequence of positive numbers and for any natural number $n$ we have
$$x_{1}+\frac{x_{4}}{2}+\frac{x_{9}}{3}+\cdots+\frac{x_{n}^{2}}{n} \leqslant 1 .$$
Prove that for any natural number $n$ we have
$$x_{1}+\frac{x_{2}}{2}+\frac{x_{3}}{3}+\cdots+\frac{x_{n}}{n}<3 ... | [Proof] Since $x_{1}, x_{2}, x_{3}, \cdots$ is a decreasing sequence of positive numbers, for any natural number $k$ we have
$$\frac{x_{k}^{2}}{k^{2}}+\frac{x_{k^{2}+1}^{2}}{k^{2}+1}+\cdots+\frac{x_{(k+1)^{2}-1}^{2}}{(k+1)^{2}-1}<(2 k+1) \frac{x_{k}^{2}}{k^{2}} \leqslant 3 \frac{x_{k}^{2}}{k} .$$
For any natural numbe... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,897 |
1. The integer 9 can be expressed as the sum of two consecutive integers $9=4+5$, and at the same time, it can be written as the sum of consecutive integers in exactly two different ways:
$$9=4+5=2+3+4 \text {. }$$
Is there an integer that can be expressed as the sum of 1990 consecutive integers, and can be written as... | [Proof] We prove that the number $m=5^{10} \cdot 199^{180}$.
(a) can be written as the sum of 1990 consecutive integers;
(b) has exactly 1990 different ways to be written as the sum of consecutive integers.
(a) In fact, let $n=\frac{5^{9} \cdot 199^{179}-1989}{2}$, then
$$\begin{aligned}
& n+(n+1)+\cdots+(n+1989) \\
= ... | proof | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 734,898 |
8・154 Given positive integers $a_{0}, a_{1}, a_{2}, \cdots, a_{99}, a_{100}$ satisfying $a_{1}>a_{0}, a_{n}=3 a_{n-1}-2 a_{n-2}, n=2,3, \cdots, 100$.
Prove: $a_{100}>2^{99}$. | ［Proof］ From the assumption, we have
$$a_{n}-a_{n-1}=2\left(a_{n-1}-a_{n-2}\right), n=2,3, \cdots, 100$$
Therefore, $a_{100}=a_{99}+2^{99}\left(a_{1}-a_{0}\right)$
Since $a_{1}, a_{0}$ are positive integers and $a_{1}>a_{0}$, it follows that $a_{1}-a_{0} \geqslant 1$. Also, $a_{99}$ is a positive integer, so
$$a_{100}... | proof | Algebra | proof | Yes | Yes | inequalities | false | 734,899 |
8-155 Let real numbers $a_{0}, a_{1}, \cdots, a_{n}$ satisfy
$$a_{0}=a_{n}=0, a_{k-1}-2 a_{k}+a_{k+1} \geqslant 0, k=1,2, \cdots, n-1 .$$
Prove: $a_{k} \leqslant 0, k=0,1,2, \cdots, n$. | [Proof] By induction. When $k=0$, the inequality to be proved is obviously true. Suppose for some $0 \leqslant k \leqslant n-2$ we have $a_{k} \leqslant 0$. From the given condition, we know
$$a_{k+2}-a_{k+1} \geqslant a_{k+1}-a_{k} \geqslant a_{k+1}$$
Thus,
$$a_{k+2} \geqslant 2 a_{k+1}$$
If $k=n-2$, then $a_{k+2}=a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,900 |
8・156 Let the sequence $a_{1}, a_{2}, a_{3}, \cdots$ satisfy
$$a_{1}=1, a_{n}=a_{n-1}+\frac{1}{a_{n-1}}, n=2,3, \cdots$$
Prove: $a_{100}>14$. | [Proof] Clearly, $a_{1}, a_{2}, a_{3}, \cdots$ are all positive numbers and
$$a_{n}^{2}=a_{n-1}^{2}+2+\frac{1}{a_{n-1}^{2}}>a_{n-1}^{2}+2, n=2,3, \cdots$$
By induction, we can obtain $a_{100}^{2}>a_{1}^{2}+2 \times 99=199$, so $a_{100}>\sqrt{199}>14$. | a_{100}>\sqrt{199}>14 | Algebra | proof | Yes | Yes | inequalities | false | 734,901 |
8-157 Given real numbers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy
$$a_{1}=0,\left|a_{k}\right|=\left|a_{k-1}+1\right|, k=2,3, \cdots, n .$$
Prove: $\frac{1}{n}\left(a_{1}+a_{2}+\cdots+a_{n}\right) \geqslant-\frac{1}{2}$. | [Proof] From the assumption, we have
$$a_{k}^{2}=a_{k-1}^{2}+2 a_{k-1}+1, \quad k=2,3, \cdots, n$$
Thus, $\quad \sum_{k=2}^{n} a_{k}^{2}=\sum_{k=2}^{n} a_{k-1}^{2}+2 \sum_{k=2}^{n} a_{k-1}+n-1$.
Since $a_{1}=0$, we have
$$2 \sum_{k=1}^{n-1} a_{k}=a_{n}^{2}+1-n$$
From this, we get $\frac{1}{n}\left(a_{1}+a_{2}+\cdots+... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,902 |
8・158 Let the infinite sequence $\left\{x_{n}\right\}$ satisfy
$$x_{0}=1,0<x_{n+1} \leqslant x_{n}, n=0,1,2, \cdots$$
(1) Prove that there exists $n \geqslant 1$ such that
$$\frac{x_{0}^{2}}{x_{1}}+\frac{x_{1}^{2}}{x_{2}}+\cdots+\frac{x_{n-1}^{2}}{x_{n}} \geqslant 3.999$$
(2) Construct a sequence that satisfies the giv... | [Solution] (1) Let $s_{n}=\frac{x_{0}^{2}}{x_{1}}+\frac{x_{1}^{2}}{x_{2}}+\cdots+\frac{x_{n-1}^{2}}{x_{n}}$, since $\left\{x_{k}\right\}$ is a non-increasing positive sequence, we have
$$\begin{aligned}
S_{2}= & \frac{x_{0}^{2}}{x_{1}}+\frac{x_{1}^{2}}{x_{2}} \geqslant 2 x_{0} \sqrt{\frac{x_{1}}{x_{2}}} \geqslant 2 x_{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,903 |
8・160 Let the sequences $\left\{x_{n}\right\}$ and $\left\{y_{n}\right\}$ satisfy
$$\begin{array}{l}
x_{0}=1, x_{1}=1, x_{n+1}=x_{n}+2 x_{n-1}, n=1,2,3, \cdots \\
y_{0}=1, y_{1}=7, y_{n+1}=2 y_{n}+3 y_{n-1}, n=1,2,3, \cdots
\end{array}$$
Prove that for any positive integers $m, n$ we have $x_{m} \neq y_{n}$. | ［Proof］It is easy to see that for $n=0,1,2,3, \cdots$ we have
$$\begin{array}{c}
x_{n}=\frac{1}{3}\left(2^{n+1}+(-1)^{n}\right) \\
y_{n}=2 \cdot 3^{n}-(-1)^{n}
\end{array}$$
If there exist positive integers $m, n$ such that $x_{m}=y_{n}$, then
$$\frac{1}{3}\left(2^{m+1}+(-1)^{m}\right)=2 \cdot 3^{n}-(-1)^{n}$$
which ... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 734,905 |
8・161 Let $\left\{a_{k}\right\}, k=1,2 \cdots$ be a sequence of non-negative real numbers, satisfying
$$a_{k}-2 a_{k+1}+a_{k+2} \geqslant 0$$
and $\quad \sum_{i=1}^{k} a_{i} \leqslant 1, k=1,2, \cdots$
Prove: For any natural number $k$,
$$0 \leqslant a_{k}-a_{k+1}<\frac{2}{k^{2}}$$ | [Proof] First, prove $0 \leqslant a_{k}-a_{k+1}$. Suppose there exists a natural number $k$ such that $a_{k}-a_{k+1}<0$, then by the assumption we have
$$a_{k+1}-a_{k+2} \leqslant a_{k}-a_{k+1}<0$$
By induction, the sequence is strictly increasing from the $k$-th term onwards, thus when $n \rightarrow+\infty$, $\sum_{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,906 |
8-162 Given the sequence $\left\{a_{n}\right\}$ satisfies
$$a_{1}=\frac{1}{2}, a_{n}=\left(\frac{2 n-3}{2 n}\right) a_{n-1}, n=2,3,4, \cdots$$
Prove: For any natural number $n$, $\sum_{k=1}^{n} a_{k}<1$. | [Proof] From the assumption, we have
$$3 a_{k-1}=2\left(k a_{k-1}-k a_{k}\right), k=2,3,4, \cdots$$
Thus, $3 \sum_{k=2}^{n+1} a_{k-1}=2 \sum_{k=2}^{n+1} k a_{k-1}-2 \sum_{k=2}^{n+1} k a_{k}$
$$\begin{array}{l}
=2 \sum_{k=1}^{n}(k+1) a_{k}-2 \sum_{k=2}^{n} k a_{k}-2(n+1) a_{n+1} \\
=2 a_{1}+2 \sum_{k=1}^{n} a_{k}-2(n+1... | proof | Algebra | proof | Yes | Yes | inequalities | false | 734,907 |
8・163 Let $a_{n}=\left[\sqrt{(n-1)^{2}+n^{2}}\right], n=1,2,3, \cdots$ where $[x]$ denotes the greatest integer not exceeding $x$. Prove:
(i) There are infinitely many positive integers $m$ such that $a_{m+1}-a_{m}>1$.
(ii) There are infinitely many positive integers $m$ such that $a_{m+1}-a_{m}=1$. | ［Proof］Obviously, for any natural number $n$ we have
$$\sqrt{2} n-32 a_{n}+1 .$$
Thus, we have
$$\begin{aligned}
a_{n+1} & =\left[\sqrt{n^{2}+(n+1)^{2}}\right] \geqslant\left[\sqrt{a_{n}^{2}+4 n}\right] \\
& \geqslant\left[\sqrt{a_{n}^{2}+2 a_{n}+1}\right]=a_{n}+1
\end{aligned}$$
That is, $\square$
$$a_{n+1} \geqslan... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 734,908 |
8-164 Let the sequence of positive numbers $\left\{a_{n}\right\}$ satisfy that for any natural numbers $k, n, m, l$, if $k+n$ $=m+l$, then
$$\frac{a_{k}+a_{n}}{1+a_{k} a_{n}}=\frac{a_{m}+a_{l}}{1+a_{m} a_{l}} .$$
Prove: There exist positive numbers $b$ and $c$, such that for any $n$ we have $b \leqslant a_{n} \leqslan... | [Proof] For $n \geqslant 2$, let
$$A_{n}=\frac{a_{1}+a_{n-1}}{1+a_{1} a_{n-1}}$$
Let $t=\min \left\{a_{1}, \frac{1}{a_{1}}\right\}$. Then, from $t \leqslant a_{1}, t a_{1} \leqslant 1$, we have
$$\begin{array}{l}
a_{1}+a_{n-1} \geqslant t+a_{1} t a_{n-1}, \text { i.e., } \\
A_{n}=\frac{a_{1}+a_{n-1}}{1+a_{1} a_{n-1}} ... | proof | Algebra | proof | Yes | Yes | inequalities | false | 734,909 |
8・165 Given a sequence of natural numbers $\left\{x_{n}\right\}$ that satisfies
$$x_{1}=a, x_{2}=b, x_{n+2}=x_{n}+x_{n+1}, n=1,2,3, \cdots$$
If one of the terms in the sequence is 1000, what is the smallest possible value of $a+b$? | [Solution] Take the sequence $\left\{t_{n}\right\}$ satisfying:
$$t_{1}=1, t_{2}=0, t_{n+2}=t_{n}+t_{n+1}, n=1,2,3, \cdots$$
It is easy to prove that for any natural number $n$ we have
$$x_{n}=t_{n} a+t_{n+1} b$$
The terms of the sequence $\left\{t_{n}\right\}$ that are less than 1000 are $1,0,1,1,2,3,5,8,13,21,34,55... | 10 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 734,910 |
8-166 Let the sequence of real numbers $a_{1}, a_{2}, a_{3}, \cdots$ satisfy: for any natural number $k$,
$$a_{k+1}=\frac{k a_{k}+1}{k-a_{k}}$$
Prove that the sequence $\left\{a_{n}\right\}$ has infinitely many positive terms and infinitely many negative terms. | [Proof] If $a_{k}=0$, then $a_{k+1}=\frac{1}{k}$. Therefore, it suffices to prove: For any natural number $N$, $a_{N}, a_{N+1}, a_{N+2}, \cdots$ cannot all be non-positive, nor can they all be non-negative.
Assume there exists a natural number $N$ such that
$a_{k} \leqslant 0$, for any $k \geqslant N$.
Since $a_{k+1}=... | proof | Algebra | proof | Yes | Yes | inequalities | false | 734,911 |
8-168 The sequence $a_{1}, a_{2}, a_{3}, \cdots$ satisfies:
$$a_{1}=1, a_{n+1}=\sqrt{a_{n}^{2}+\frac{1}{a_{n}}}, n=1,2,3, \cdots$$
Prove that there exists a positive number $\alpha$, such that $\frac{1}{2} \leqslant \frac{a_{n}}{n^{\alpha}} \leqslant 2, n=1,2,3, \cdots$ | [Proof] By induction, we can prove that
$$\frac{1}{2} n^{\frac{1}{3}} \leqslant a_{n} \leqslant 2 n^{\frac{1}{3}}, n=1,2, \cdots$$
i.e., $\alpha=-\frac{1}{3}$ is the desired value. In fact, when $n=1$, by $a_{1}=1$, (1) clearly holds. Suppose when $n=k$, (1) holds. Then,
$$a_{k+1}=\sqrt{a_{k}^{2}+\frac{1}{a_{k}}} \leq... | proof | Algebra | proof | Yes | Yes | inequalities | false | 734,913 |
8- 169 Given the sequence $\left\{a_{n}\right\}$, let $a_{0}=0$. Define
$$b_{n}=a_{n}-a_{n-1}, c_{n}=\frac{a_{n}}{n}, n=1,2,3, \cdots$$
If the sequence $\left\{b_{n}\right\}$ is monotonically decreasing, prove that the sequence $\left\{c_{n}\right\}$ is also monotonically decreasing. | [Proof] Since $a_{0}=0, b_{k}=a_{k}-a_{k-1}(k \geqslant 1)$, then
$$\begin{aligned}
a_{n}= & a_{0}+\left(a_{1}-a_{0}\right)+\left(a_{2}-a_{1}\right)+\cdots+\left(a_{n}-a_{n-1}\right)=b_{1}+b_{2}+ \\
& \cdots b_{n} .
\end{aligned}$$
By $\left\{b_{n}\right\}$ being monotonically decreasing, we have
$$a_{n} \geqslant n b... | proof | Algebra | proof | Yes | Yes | inequalities | false | 734,914 |
8. 170 Prove: There exists a positive number $k$, such that for any positive sequence $a_{1}, a_{2}, a_{3}, \cdots$ if $\sum_{n=1}^{\infty} \frac{1}{a_{n}}<+\infty$, then
$$\sum_{n=1}^{\infty} \frac{n}{a_{1}+a_{2}+\cdots+a_{n}} \leqslant k \sum_{n=1}^{\infty} \frac{1}{a_{n}}$$ | [Proof] For any even number $m=2l$, it can be proven that
$$\sum_{n=1}^{m} \frac{n}{a_{1}+a_{2}+\cdots+a_{n}} \leqslant 4 \sum_{n=1}^{m} \frac{1}{a_{n}}$$
In fact, let $b_{1}, b_{2}, \cdots, b_{m}$ be a permutation of $a_{1}, a_{2}, \cdots, a_{m}$, and $b_{1} \leqslant b_{2} \leqslant \cdots \leqslant b_{m}$. For any ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,915 |
8. 172 Let $a_{1}, a_{2}, a_{3}, \cdots$ be a sequence of positive numbers. Prove that there exist infinitely many $n \in$ $N$, such that
$$\frac{a_{1}+a_{n+1}}{a_{n}}>1+\frac{1}{n}$$ | [Proof] By contradiction. Suppose the conclusion to be proved does not hold, then there exists $k \in \mathbb{N}$, such that when $n \geqslant k$ we have
$$\frac{a_{1}+a_{n+1}}{a_{n}} \leqslant 1+\frac{1}{n} \text {, }$$
which implies $\frac{a_{n}}{n} \geqslant \frac{a_{1}}{n+1}+\frac{a_{n+1}}{n+1}$.
Thus, we can obta... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,916 |
8. 173 Find all $a_{0} \in R$, such that the sequence
$$a_{n+1}=2^{n}-3 a_{n}, n=0,1,2, \cdots$$
determined by this is increasing. | [Solution] Since for $n \geqslant 0$ we have
$$\begin{aligned}
a_{n+1} & =2^{n}-3 a_{n}=2^{n}-3 \cdot 2^{n-1}+3^{2} a_{n-1} \\
& =2^{n}-3 \cdot 2^{n-1}+3^{2} \cdot 2^{n-2}-3^{3} a_{n-2} \\
& =\cdots=\sum_{k=0}^{n}(-1)^{k} 2^{n-k} \cdot 3^{k}+(-1)^{n+1} 3^{n+1} a_{0} \\
& =\frac{1}{5}\left(2^{n+1}-(-1)^{n+1} 3^{n+1}\rig... | a_{0}=\frac{1}{5} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 734,917 |
8.174 Given a real number $a_{1}$, let the sequence $a_{1}, a_{2}, \cdots$ satisfy
$$a_{n+1}=\left\{\begin{array}{ll}
\frac{1}{2}\left(a_{n}-\frac{1}{a_{n}}\right), & \text { when } a_{n} \neq 0, \\
0, & \text { when } a_{n}=0 .
\end{array}\right.$$
Prove: This sequence contains infinitely many non-positive terms. | [Proof] By contradiction. Suppose the sequence has only a finite number of non-positive terms, then there exists $n_{0} \in \mathbb{N}$, such that when $n \geqslant n_{0}$, $a_{n}>0$. From $a_{n+1}=\frac{1}{2}\left(a_{n}-\frac{1}{a_{n}}\right)>0$ we get $a_{n}>1$, for any $n \geqslant n_{0}$. On the other hand, since
$... | proof | Algebra | proof | Yes | Yes | inequalities | false | 734,919 |
8-175 Let the sequence $\left\{a_{n}\right\}$ satisfy
$\left|a_{k+m}-a_{k}-a_{m}\right| \leqslant 1$, for any $k, m \in N$. Prove: For any $p, q \in N$,
$$\left|\frac{a_{p}}{p}-\frac{a_{q}}{q}\right|<\frac{1}{p}+\frac{1}{q} .$$ | [Proof] Let $k, m \in \mathbb{N}, k \geqslant 2$, then
$$\begin{aligned}
\left|a_{k m}-k a_{m}\right| & =\left|\sum_{i=1}^{k-1}\left(a_{(i+1) m}-a_{i m}-a_{m}\right)\right| \\
& \leqslant \sum_{i=1}^{k-1}\left|a_{(i+1) m}-a_{i m}-a_{m}\right|
\end{aligned}$$
By the assumption, we have
$$\left|a_{k m}-k a_{m}\right| \l... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,920 |
8. 177 Given $A>1, B>1$ and a sequence $\left\{a_{n}\right\}$ consisting of numbers from the interval $[1, A B]$. Prove: There exists a sequence $\left\{b_{n}\right\}$ consisting of numbers from the interval $[1, A]$, such that for any $m, n \in N$, we have
$$\frac{a_{m}}{a_{n}} \leqslant B \frac{b_{m}}{b_{n}}$$ | [Proof] Take
$$b_{n}=\left\{\begin{array}{ll}
a_{n}, & \text { when } a_{n} \leqslant A, \\
A, & \text { when } a_{n}>A.
\end{array}\right.$$
Let $C_{n}=\frac{a_{n}}{b_{n}}$, clearly $C_{n} \geqslant 1$, and since $1 \leqslant a_{n} \leqslant A B$, we have $C_{n} \leqslant B$. Therefore, for any $m, n \in N$,
$$\frac{... | proof | Other | proof | Yes | Yes | inequalities | false | 734,922 |
8. 178 The sequence $\left\{F_{n}\right\}$ is defined as follows: $F_{1}=1, F_{2}=2$, and
$$F_{n+2}=F_{n+1}+F_{n}, n=1,2,3, \cdots$$
Prove that for any natural number $n$, we have
$$\sqrt[n]{F_{n+1}} \geqslant 1+\frac{1}{\sqrt[n]{F_{n}}}$$ | [Proof] Let $F_{0}=1$, then
$$F_{k+1}=F_{k}+F_{k-1}$$
i.e. $\square$
$$1=\frac{F_{k}}{F_{k+1}}+\frac{F_{k-1}}{F_{k+1}}, k=1,2,3, \cdots$$
Thus,
$$n=\sum_{k=1}^{n} \frac{F_{k}}{F_{k+1}}+\sum_{k=1}^{n} \frac{F_{k-1}}{F_{k+1}}$$
By the AM-GM inequality, we have
$$1 \geqslant \sqrt[n]{\frac{F_{1}}{F_{2}} \cdot \frac{F_{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,923 |
8- 179 Let the sequence of real numbers $\left\{x_{n}\right\}$ satisfy: $0 \leqslant x_{1}<1$, and for $n \geqslant 1$ we have
$$x_{n+1}=\left\{\begin{array}{l}
0, \text { when } x_{n}=0 \text {, } \\
\frac{1}{x_{n}}-\left[\frac{1}{x_{n}}\right], \text { when } x_{n} \neq 0 \text {, }
\end{array}\right.$$
Prove that f... | [Proof] Let $f(x)=\frac{1}{1+x}$. Clearly, for $0 \leqslant x<y$ we have
$$0<f(x)-f(y)=\frac{y-x}{(1+x)(1+y)}<y-x$$
From this, it is easy to prove that for any natural number $n$, when $x \geqslant 0$,
$$g_{n}(x)=x+f(x)+f(f(x))+\cdots+f^{(n)}(x) \text { is strictly increasing, }$$
where $f^{(n)}(x)=f(f(\cdots f(x) \c... | proof | Other | proof | Yes | Yes | inequalities | false | 734,924 |
8. 180 In a finite sequence of real numbers $a_{1}, a_{2}, \cdots, a_{n}$, if the arithmetic mean of a segment $a_{k}, a_{k+1}, \cdots, a_{k+1}$ is greater than 1988, then we call this segment a "dragon," and $a_{k}$ is called the "dragon head" (if a single term $a_{m}>1988$, then this single term is also a dragon, and... | [Proof] In the sequence $a_{1}, a_{2}, \cdots, a_{n}$, from front to back, first take a dragon with the first dragon head as its head. Then, from the remaining terms, take a dragon with the first term that can be a dragon head as its head. Continue this process until all terms are taken. In this way, a finite number of... | proof | Algebra | proof | Yes | Yes | inequalities | false | 734,925 |
8・182 Let the sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ satisfy
$$\begin{array}{l}
a_{0}=\frac{\sqrt{2}}{2}, \quad a_{n+1}=\frac{\sqrt{2}}{2} \sqrt{1-\sqrt{1-a_{n}^{2}}}, n=0,1,2, \cdots \\
b_{0}=1, b_{n+1}=\frac{\sqrt{1+b_{n}^{2}}-1}{b_{n}}, n=0,1,2, \cdots
\end{array}$$
Prove that for any non-negat... | [Proof] $a_{0}=\frac{\sqrt{2}}{2}=\sin \frac{\pi}{4}$, if $a_{n}=\sin \frac{\pi}{2^{n+2}}$, then $a_{n+1}=\frac{\sqrt{2}}{2} \sqrt{1-\cos \frac{\pi}{2^{n+2}}}=\sin \frac{\pi}{2^{n+3}}$.
Thus, by induction, we can prove
$$a_{n}=\sin \frac{\pi}{2^{n+2}}, n=0,1,2, \cdots$$
Similarly, we can prove
$$b_{n}=\operatorname{tg... | proof | Algebra | proof | Yes | Yes | inequalities | false | 734,926 |
$1 \cdot 13$ Given a finite sequence of 0s and 1s, it has the property:
（1）If 5 consecutive digits are extracted from somewhere in the sequence and 5 consecutive digits are also extracted from any other place, then these 5 digits will not be the same (they can overlap, for example 0110101).
（2）If the digit 0 or 1 is ad... | [Proof] Let $a b c d$ be the last four digits. In the sequence, there must be five consecutive digits $a b c d 0$ (otherwise, we could add 0 and still maintain property (1), which would contradict property (2)) and $a b c d 1$. Therefore, the quadruplet $a b c d$ appears three times in the sequence.
Since 0 or 1 can a... | proof | Combinatorics | proof | Yes | Yes | inequalities | false | 734,927 |
$8 \cdot 184$ Let $1<x_{1}<2$, and define
$$x_{n+1}=1+x_{n}-\frac{1}{2} x_{n}^{2}, n=1,2,3, \cdots$$
Prove: $\left|x_{n}-\sqrt{2}\right|<2^{-n}, n=3,4,5, \cdots$ | [Proof] From the assumption, we have
$$x_{n+1}=-\frac{1}{2}\left(x_{n}-1\right)^{2}+\frac{3}{2}, n=1,2,3, \cdots$$
Since $1-\frac{1}{2}\left(\sqrt{2}+2^{-k}-1\right)^{2}+\frac{3}{2} \\
& =\sqrt{2}-\sqrt{2} \cdot 2^{-k}+2^{-k}-2^{-2 k-1}
\end{aligned}$$
From $k \geqslant 3$, we get
$$\sqrt{2}-1+2^{-k-1}\sqrt{2}-2^{-k-... | proof | Algebra | proof | Yes | Yes | inequalities | false | 734,929 |
8-185 Let the sequence of non-negative numbers $a_{1}, a_{2}, \cdots$ satisfy the condition
$$a_{n+m} \leqslant a_{n}+a_{m}, m, n \in N$$
Prove that for any $n \geqslant m$,
$$a_{n} \leqslant m a_{1}+\left(\frac{n}{m}-1\right) a_{m}$$ | [Proof] Let $n=m q+r, \quad q \in \mathbb{N}, 0 \leqslant r<m$. Then, we have
$$\begin{array}{l}
a_{n} \leqslant a_{m q}+a_{r} \\
\leqslant q a_{m}+a_{r} \\
=\frac{n-r}{m} \cdot a_{m}+a_{r} \\
=\left(\frac{n}{m}-1\right) a_{m}+\frac{m-r}{m} a_{m}+a_{r} \\
\leqslant\left(\frac{n}{m}-1\right) a_{m}+\frac{m-r}{m} \cdot m... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,930 |
8・186 $\left\{a_{n} \mid n \in N\right\}$ is a sequence of positive integers such that for each $n$,
$$\left(a_{n}-1\right)\left(a_{n}-2\right) \cdots\left(a_{n}-n^{2}\right)$$
is a positive integer and a multiple of $n^{n^{2}-1}$. Prove that for any finite set $P$ whose elements are all prime numbers, the following i... | [Proof] If $n=p$, where $p$ is any prime number, then, by the given,
$$\left(a_{p}-1\right)\left(a_{p}-2\right) \cdots\left(a_{p}-p^{2}\right)$$
is a positive integer and is a multiple of $p^{p^{2}-1}$. Note that $a_{p}>0, a_{p}-1, a_{p}-2, \cdots, a_{p} - p^{2}$ are $p^{2}$ consecutive integers, none of which are zer... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 734,931 |
9. 1 For which values of $x$ is the inequality
$$\frac{4 x^{2}}{(1-\sqrt{1+2 x})^{2}}<2 x+9$$
true. | [Solution] From the assumption, we know that $1+2 x \geqslant 0,1-\sqrt{1+2 x} \neq 0$, so $x \geqslant-\frac{1}{2}$ and $x \neq 0$.
Also, when $x \neq 0$,
$$\frac{-2 x}{1-\sqrt{1+2 x}}=1+\sqrt{1+2 x}$$
Thus, the original inequality becomes
$$2+2 x+2 \sqrt{1+2 x}<2 x+9$$
Solving this, we get $-\frac{1}{2} \leqslant x... | -\frac{1}{2} \leqslant x < 0 \text{ or } 0 < x < \frac{45}{8} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 734,932 |
9.2 Find all real numbers $x$ that satisfy the inequality
$$\sqrt{3-x}-\sqrt{x+1}>\frac{1}{2}$$ | [Solution] From $3-x \geqslant 0, x+1 \geqslant 0, \sqrt{3-x}>\sqrt{x+1}$, we know
$$-1 \leqslant x<1+\frac{\sqrt{31}}{8}$$.
From (1) and (2), the solution to the original inequality is
$$-1 \leqslant x<1-\frac{\sqrt{31}}{8}$$ | -1 \leqslant x<1-\frac{\sqrt{31}}{8} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 734,933 |
$9 \cdot 3$ Find all real numbers $x$ that satisfy the following conditions:
$$0 \leqslant x \leqslant 2 \pi \text { and } 2 \cos x \leqslant|\sqrt{1+\sin 2 x}-\sqrt{1-\sin 2 x}| \leqslant \sqrt{2} \text {. }$$ | [Solution] Clearly, for any real number $x$ we have
$$|\sqrt{1+\sin 2 x}-\sqrt{1-\sin 2 x}| \leqslant \sqrt{2}$$
The problem reduces to solving the inequality in $[0,2 \pi]$
$$2 \cos x \leqslant|\sqrt{1+\sin 2 x}-\sqrt{1-\sin 2 x}|$$
When $\frac{\pi}{2} \leqslant x \leqslant \frac{3 \pi}{2}$, since $\cos x \leqslant ... | \frac{\pi}{4} \leqslant x \leqslant \frac{7 \pi}{4} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 734,934 |
9・4 Find all pairs of numbers $(x, y)$ that satisfy
$$|\sin x-\sin y|+\sin x \cdot \sin y \leqslant 0$$ | [Solution] If $(x, y)$ satisfies the given inequality, then $\sin x \cdot \sin y \leqslant 0$,
from which we can get $|\sin x-\sin y|=|\sin x|+|\sin y|$.
Thus, $(x, y)$ satisfies $|\sin x|+|\sin y|+\sin x \cdot \sin y \leqslant 0$, hence $\sin x=\sin y=0$.
The set of all pairs $(x, y)$ is
$$\{(x, y) \mid x=n \pi, y=m \... | \{(x, y) \mid x=n \pi, y=m \pi, n, m \in \mathbb{Z}\} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 734,935 |
$9 \cdot 5$ Given $f(x)=x^{2}-6 x+5$, find the range of points $(x, y)$ on the plane that satisfy
$f(x)+f(y) \leqslant 0$ and $f(x)-f(y) \geqslant 0$. And plot the graph. | [Solution] Since $f(x)+f(y)=x^{2}+y^{2}-6 x-6 y+10$
$$=(x-3)^{2}+(y-3)^{2}-8$$
Therefore, the points $(x, y)$ that satisfy $f(x)+f(y) \leqslant 0$ lie on or inside the circle $(x-3)^{2}+(y-3)^{2}=8$.
From $f(x)-f(y)=x^{2}-y^{2}-6 x+6 y$ $=(x-y)(x+y-6)$, we know that $f(x)-f(y) \geqslant 0$ is equivalent to
$$\begin{a... | not found | Algebra | math-word-problem | Yes | Yes | inequalities | false | 734,936 |
$9 \cdot 6$ Calculate the following product, accurate to within 0.00001:
$$\left(1-\frac{1}{10}\right)\left(1-\frac{1}{10^{2}}\right) \cdots\left(1-\frac{1}{10^{99}}\right) .$$ | [Solution] Let $x=\left(1-\frac{1}{10}\right)\left(1-\frac{1}{10^{2}}\right) \cdots\left(1-\frac{1}{10^{99}}\right)$. Since $1-\frac{1}{9 \cdot 10^{5}}<1-\left(\frac{1}{10^{6}}+\cdots+\frac{1}{10^{99}}\right)<\left(1-\frac{1}{10^{6}}\right) \cdots\left(1-\frac{1}{10^{99}}\right)<$ 1, it follows that $0<\left(1-\frac{1}... | 0.89001 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 734,937 |
9.7 Try to select 100 numbers such that they satisfy
$$x_{1}=1,0 \leqslant x_{k} \leqslant 2 x_{k-1}, k=2,3, \cdots, 100$$
and make $s=x_{1}-x_{2}+x_{3}-x_{4}+\cdots+x_{99}-x_{100}$ as large as possible | ［Solution］ From the assumption, we have
$$\begin{aligned}
s & \leqslant x_{1}+\left(-x_{2}+x_{3}\right)+\left(-x_{4}+x_{5}\right)+\cdots+\left(-x_{98}+x_{99}\right) \\
& \leqslant x_{1}+x_{2}+x_{4}+\cdots+x_{98} \\
& \leqslant 1+2+2^{3}+\cdots+2^{97}
\end{aligned}$$
From this, we know that $x_{1}=1, x_{2}=2, x_{3}=2^{... | 1+2+2^{3}+\cdots+2^{97} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 734,938 |
1. 125 Given 33 positive integers, all of whose prime factors are $2,3,5,7$ and 11. Prove: Among these 33 positive integers, there must be two whose product is a perfect square. | [Proof] Let these 33 positive integers be $n_{1}, n_{2}, \cdots, n_{33}$.
From the problem, we know that $n_{k}=2^{a_{k}} 3^{b_{k}} 5^{c_{k}} 7^{d_{k}} 11^{e_{k}}$, where $k=1,2, \cdots, 33 ; a_{k}, b_{k}, c_{k}, d_{k}, e_{k}$ are all non-negative integers.
Clearly, $n_{i} n_{k}$ is a perfect square if and only if $1 ... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 734,939 |
9.8 Find the region composed of $(x, y)$ that satisfies the inequality $\log _{x} y \geqslant \log _{y} \frac{x}{y}(x y)$. | ［Solution］ Clearly, we have $x>0, y>0, x \neq 1, x \neq y$. Let
$$u=\log _{x} y$$
Then $u \neq 1$, and
$$\log _{y}(x y)=\frac{\log _{x}(x y)}{\log _{x} \frac{x}{y}}=\frac{1+u}{1-u}$$
Clearly, $u \neq \frac{1+u}{1-u}$, and
$$u>\frac{1+u}{1-u} \Leftrightarrow u-\frac{1+u}{1-u}=\frac{u^{2}+1}{u-1}>0 \Leftrightarrow u>1$... | \{(x, y) \mid 01 \text { and } y>x\} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 734,940 |
9.9 Prove: For any real numbers $x, y, z$, the following three inequalities cannot all hold simultaneously:
$|x|<|y-z|,|y|<|z-x|,|z|<|x-y|$. | [Proof] By contradiction, assume the three inequalities hold simultaneously, then
$$\left\{\begin{array}{l}
x^{2}<(y-z)^{2} \\
y^{2}<(z-x)^{2} \\
z^{2}<(x-y)^{2}
\end{array}\right.$$
Thus, we can obtain
$$\left\{\begin{array}{l}
(x-y+z)(x+y-z)<0 \\
(y-z+x)(y+z-x)<0 \\
(z-x+y)(z+x-y)<0
\end{array}\right.$$
Multiplying... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,941 |
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