2002 AIME II 第 6 题

Question

求与下式最接近的整数：

1000 \sum_{n=3}^{10000} \frac{1}{n^2 - 4}

Find the integer that is closest to

1000 \sum_{n=3}^{10000} \frac{1}{n^2 - 4}.

答案

分解

n^2 - 4 = (n-2)(n+2)

，并用部分分式让求和裂项相消。

Factor

n^2 - 4 = (n-2)(n+2)

and use partial fractions to make the sum telescope

前端只剩下

1

、

\frac{1}{2}

、

\frac{1}{3}

、

\frac{1}{4}

，再减去四个接近

\frac{1}{10000}

的小尾项。

Only

1,

\frac{1}{2},

\frac{1}{3},

\frac{1}{4}

survive at the front, minus four tiny tail terms near

\frac{1}{10000}

因为

\frac{1}{n^2 - 4} = \frac{1}{4}\left(\frac{1}{n-2} - \frac{1}{n+2}\right)

，所以该和裂项相消：

\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}\text{。}

令

r

等于

250

乘以四个正尾项之和。于是

0 \lt r \lt 250 \cdot \frac{4}{9999} \lt \frac{1}{3}

。原式的值为

520 + \frac{5}{6} - r

，所以它严格介于

520 + \frac{1}{2}

与

520 + \frac{5}{6}

之间。因此最接近的整数是

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}

Let

r

be

250

times the sum of the four positive tail fractions. Then

0 \lt r \lt 250 \cdot \frac{4}{9999} \lt \frac{1}{3}.

The value of the expression is

520 + \frac{5}{6} - r,

so it lies strictly between

520 + \frac{1}{2}

and

520 + \frac{5}{6}.

Hence the closest integer is

521.

其他年份的第 6 题