Question

考虑由 $a_k = \frac{1}{k^2 + k}$ （ $k \ge 1$ ）定义的数列。已知正整数 $m$ 和 $n$ 满足 $m \lt n$ ，且 $a_m + a_{m+1} + \cdots + a_{n-1} = \frac{1}{29}$ 。求 $m + n$ 。

Consider the sequence defined by $a_k = \frac{1}{k^2 + k}$ for $k \ge 1.$ Given that $a_m + a_{m+1} + \cdots + a_{n-1} = \frac{1}{29},$ for positive integers $m$ and $n$ with $m \lt n,$ find $m + n.$

答案

答案：840

知识点：裂项相消部分分式丢番图方程

难度评级：2110

解答：

因为 $\frac{1}{k^2 + k} = \frac{1}{k(k + 1)} = \frac{1}{k} - \frac{1}{k + 1}$ ，所以该和望远镜相消： $\begin{aligned} &a_m + a_{m+1} + \cdots + a_{n-1} \\ &= \frac{1}{m} - \frac{1}{n} \\ &= \frac{1}{29}. \end{aligned}$

两边乘以 $29mn$ ，得 $29n - 29m = mn$ ，整理为 $(29 - m)(29 + n) = 29^2$ 。由于 $29$ 是质数且 $29 + n \gt 29$ ，在 $m$ 为正整数时唯一的分解是 $29 - m = 1$ 、 $29 + n = 841$ ，所以 $m = 28$ 、 $n = 812$ 。

因此 $m + n = 28 + 812 = 840$ 。

Since $\frac{1}{k^2 + k} = \frac{1}{k(k + 1)} = \frac{1}{k} - \frac{1}{k + 1},$ the sum telescopes: $\begin{aligned} &a_m + a_{m+1} + \cdots + a_{n-1} \\ &= \frac{1}{m} - \frac{1}{n} \\ &= \frac{1}{29}. \end{aligned}$

Multiplying through by $29mn$ gives $29n - 29m = mn,$ which rearranges to $(29 - m)(29 + n) = 29^2.$ Since $29$ is prime and $29 + n \gt 29,$ the only factorization with $m$ a positive integer is $29 - m = 1$ and $29 + n = 841,$ so $m = 28$ and $n = 812.$

Therefore $m + n = 28 + 812 = 840.$

2002 AIME I 第 4 题

答案：840

难度评级：2110

解答：

其他年份的第 4 题