Question

求最接近 $1000 \sum_{n=3}^{10000} \frac{1}{n^2 - 4}.$ 的整数。

Find the integer that is closest to $1000 \sum_{n=3}^{10000} \frac{1}{n^2 - 4}.$

答案

答案：521

难度评级：2340

解答：

因为 $\frac{1}{n^2 - 4} = \frac{1}{4}\left(\frac{1}{n-2} - \frac{1}{n+2}\right)$ ，所以该和望远镜相消： $\begin{aligned} &1000 \sum_{n=3}^{10000} \frac{1}{n^2 - 4} \\ &\tiny{}= 250\left(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} - \frac{1}{9999} - \frac{1}{10000} - \frac{1}{10001} - \frac{1}{10002}\right). \end{aligned}$

前面的部分是 $250 \cdot \frac{25}{12} = 520.8\overline{3}$ ，四个尾项只减去大约 $250 \cdot \frac{4}{10000} = 0.1$ 。因此该值约为 $520.73$ ，最接近的整数是 $521$ 。

Since $\frac{1}{n^2 - 4} = \frac{1}{4}\left(\frac{1}{n-2} - \frac{1}{n+2}\right),$ the sum telescopes: $\begin{aligned} &1000 \sum_{n=3}^{10000} \frac{1}{n^2 - 4} \\ &\tiny{}= 250\left(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} - \frac{1}{9999} - \frac{1}{10000} - \frac{1}{10001} - \frac{1}{10002}\right). \end{aligned}$

The front part is $250 \cdot \frac{25}{12} = 520.8\overline{3},$ and the four tail terms subtract only about $250 \cdot \frac{4}{10000} = 0.1.$ The value is therefore about $520.73,$ so the closest integer is $521.$

2002 AIME II 第 6 题

答案：521

难度评级：2340

解答：

其他年份的第 6 题