2016 UNCO Math Contest II Problems/Problem 7: Difference between revisions

Revision as of 19:18, 4 March 2021

Evaluate $S =\sum_{n=2}^{\infty} \frac{4n}{(n^2-1)^2}$

$\frac{5}{4}$

This is a telescoping series:

(1−1/9)+(1/4−1/16)+(1/9−1/25)+(1/16−1/36)+(1/25−1/49)+...=5/4

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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
+(1−1/9)+(1/4−1/16)+(1/9−1/25)+(1/16−1/36)+(1/25−1/49)+...=5/4
 == 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