2002 AIME II Problems/Problem 6

Revision as of 23:37, 8 March 2014 by MSTang (talk | contribs) (Solution)

Problem

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

Solution

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

So if we 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:

$250 (1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} - \frac{1}{9999} - \frac{1}{10000} - \frac{1}{10001} - \frac{1}{10002}) = 250 + 125 + 83.3 + 62.5 - 250 (- \frac{1}{9999} - \frac{1}{10000} - \frac{1}{10001} - \frac{1}{10002})$

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


If you didn't know $\frac{4}{n^2 - 4} = \frac{1}{n-2} - \frac{1}{n + 2}$, here's how you can find it out:

We know $\frac{1}{n^2 - 4} = \frac{1}{(n+2)(n-2)}$. We can use the process of fractional decomposition to split this into two fractions thus: $\frac{1}{(n+2)(n-2)} = \frac{A}{(n+2)} + \frac{B}{(n+2)}$ for some A and B.

Solving for A and B gives $1 = (n-2)A + (n+2)B$ or $1 = n(A+B)+ 2(B-A)$. Since there is no n term on the left hand side, $A+B=0$ and by inspection $1 = 2(B-A)$. Solving yields $A=\frac{1}{4},  B=\frac{-1}{4}$

Then we have $\frac{1}{(n+2)(n-2)} = \frac{ \frac{1}{4} }{(n-2)} + \frac{ \frac{-1}{4} }{(n+2)}$ and we can continue as before.

See also

2002 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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Although the answer to Problem 6 doesn't change, the value of the telescoping sum is incorrect as given. Instead of \[250 \left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{9997}-\frac{1}{9998}-\frac{1}{9999}-\frac{1}{10000} \right),\] the correct sum is \[250 \left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{9999}-\frac{1}{10000}-\frac{1}{10001}-\frac{1}{10002} \right).\]

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