Difference between revisions of "2002 AIME II Problems/Problem 6"
I like pie (talk | contribs) (Added problem, solution still needed) |
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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 .
Solution
This problem needs a solution. If you have a solution for it, please help us out by adding it.
You know that .
So if you pull the out of the summation, you get
.
Now that telescopes, leaving you with:
is not enough to bring lower than so the answer is
See also
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 | ||
All AIME Problems and Solutions |