Difference between revisions of "2002 AIME II Problems/Problem 6"
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== Solution == | == Solution == | ||
− | + | We know that <math>\frac{4}{n^2 - 4} = \frac{1}{n-2} - \frac{1}{n + 2}</math>. | |
− | So if | + | So if we pull the <math>\frac{1}{4}</math> out of the summation, you get |
<math>250\sum_{n=3}^{10,000} (\frac{1}{n-2} - \frac{1}{n + 2})</math>. | <math>250\sum_{n=3}^{10,000} (\frac{1}{n-2} - \frac{1}{n + 2})</math>. | ||
Line 11: | Line 11: | ||
Now that telescopes, leaving you with: | Now that telescopes, leaving you with: | ||
− | <math>250 (1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} - \frac{1}{ | + | <math>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})</math> |
<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> | <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> |
Revision as of 23:37, 8 March 2014
Problem
Find the integer that is closest to .
Solution
We know that .
So if we 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
If you didn't know , here's how you can find it out:
We know . We can use the process of fractional decomposition to split this into two fractions thus: for some A and B.
Solving for A and B gives or . Since there is no n term on the left hand side, and by inspection . Solving yields
Then we have and we can continue as before.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
Although the answer to Problem 6 doesn't change, the value of the telescoping sum is incorrect as given. Instead of the correct sum is