Difference between revisions of "2002 AIME II Problems/Problem 6"

(Added problem, solution still needed)
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{{solution}}
 
{{solution}}
  
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You know that <math>\frac{4}{n^2 - 4} = \frac{1}{n-2} - \frac{1}{n + 2}</math>.
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So if you pull the <math>\frac{1}{4}</math> out of the summation, you get
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<math>250\sum_{n=3}^{10,000} (\frac{1}{n-2} - \frac{1}{n + 2})</math>.
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Now that telescopes, leaving you with:
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<math>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})</math>
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<math>250(-\frac{1}{9997} - \frac{1}{9998} - \frac{1}{9999} - \frac{1}{10000})</math> is not enough to bring <math>520.8</math> lower than <math>520.5</math>  so the answer is <math>\fbox{521}</math>
 
== See also ==
 
== See also ==
 
{{AIME box|year=2002|n=II|num-b=5|num-a=7}}
 
{{AIME box|year=2002|n=II|num-b=5|num-a=7}}

Revision as of 14:46, 23 June 2008

Problem

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

Solution

This problem needs a solution. If you have a solution for it, please help us out by adding it.

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 $\fbox{521}$

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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