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

(Solution)
(Solution 2)
 
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==Solution 2==
 
==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:
 
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 ==
 
== See also ==

Latest revision as of 20:18, 4 March 2021

Problem

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

Solution

$\frac{5}{4}$

Solution 2

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

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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