2002 AIME II Problems/Problem 6

Problem

Find the integer that is closest to $1000\sum_{n=3}^{10000}\frac1{n^2-4}$ .

Solution

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You know that $\frac{4}{n^2 - 4} = \frac{1}{n-2} - \frac{1}{n + 2}$ .

So if you pull the $\frac{1}{4}$ out of the summation, you get

$250\sum_{n=3}^{10,000} (\frac{1}{n-2} - \frac{1}{n + 2})$ .

Now that telescopes, leaving you with:

$250 (1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} - \frac{1}{9997} - \frac{1}{9998} - \frac{1}{9999} - \frac{1}{10000}) = 250 + 125 + 83.3 + 62.5 - 250 (- \frac{1}{9997} - \frac{1}{9998} - \frac{1}{9999} - \frac{1}{10000})$

$250(-\frac{1}{9997} - \frac{1}{9998} - \frac{1}{9999} - \frac{1}{10000})$ is not enough to bring $520.8$ lower than $520.5$ so the answer is $\fbox{521}$

2002 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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2002 AIME II Problems/Problem 6

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