Difference between revisions of "2016 UNCO Math Contest II Problems/Problem 7"
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== Solution == | == Solution == | ||
<math>\frac{5}{4}</math> | <math>\frac{5}{4}</math> | ||
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| + | ==Solution 2== | ||
| + | |||
| + | Rewrite the term: | ||
| + | |||
| + | (∑n=2∞4n(n2−1)2)=(∑n=2∞(−1(n+1)2+1(n−1)2)) | ||
| + | This is a telescoping series: | ||
| + | |||
| + | ∑n=2∞(−1(n+1)2+1(n−1)2)=(1−19)+(14−116)+(19−125)+(116−136)+(125−149)+...=54 | ||
== See also == | == See also == | ||
Revision as of 19:16, 4 March 2021
Contents
Problem
Evaluate
![\[S =\sum_{n=2}^{\infty} \frac{4n}{(n^2-1)^2}\]](http://latex.artofproblemsolving.com/5/8/f/58f783b714712db7d5b19d0f2b8d4de5d847cb83.png)
Solution
Solution 2
Rewrite the term:
(∑n=2∞4n(n2−1)2)=(∑n=2∞(−1(n+1)2+1(n−1)2)) This is a telescoping series:
∑n=2∞(−1(n+1)2+1(n−1)2)=(1−19)+(14−116)+(19−125)+(116−136)+(125−149)+...=54
See also
| 2016 UNCO Math Contest II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 6 |
Followed by Problem 8 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 | ||
| All UNCO Math Contest Problems and Solutions | ||
