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

(Solution)
(Solution)
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== Solution ==
 
== Solution ==
 
<math>\frac{5}{4}</math>
 
<math>\frac{5}{4}</math>
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==Solution 2==
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Rewrite the term:
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(∑n=2∞4n(n2−1)2)=(∑n=2∞(−1(n+1)2+1(n−1)2))
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This is a telescoping series:
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∑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 20:16, 4 March 2021

Problem

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

Solution

$\frac{5}{4}$

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 (ProblemsAnswer KeyResources)
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
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