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

(Solution 2)
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Evaluate
 
Evaluate
 
<cmath>S =\sum_{n=2}^{\infty} \frac{4n}{(n^2-1)^2}</cmath>
 
<cmath>S =\sum_{n=2}^{\infty} \frac{4n}{(n^2-1)^2}</cmath>
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== Solution ==
 
== Solution ==
<math>\frac{5}{4}</math>
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First, we perform fractional decomposition on the summed expression.
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Let <cmath>\frac{A}{(n-1)^2}+\frac{B}{(n+1)^2} = \frac{4n}{(n^2-1)^2}</cmath>.
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Multiplying both sides by <math>(n^2-1)^2</math> and expanding gives <math>(A+B)n^2+2(A-B)n+(A+B)=4n</math>
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Therefore, we have the system of equations <math>\begin{cases} A+B=0\
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A-B=2\end{cases}</math>, which has the solution <math>(A,B)=(1,-1)</math>.
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<math>\boxed{\frac{5}{4}}</math>
  
 
==Solution 2==
 
==Solution 2==

Revision as of 18:06, 21 October 2023

Problem

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


Solution

First, we perform fractional decomposition on the summed expression. Let \[\frac{A}{(n-1)^2}+\frac{B}{(n+1)^2} = \frac{4n}{(n^2-1)^2}\]. Multiplying both sides by $(n^2-1)^2$ and expanding gives $(A+B)n^2+2(A-B)n+(A+B)=4n$ Therefore, we have the system of equations $\begin{cases} A+B=0\\ A-B=2\end{cases}$, which has the solution $(A,B)=(1,-1)$.

$\boxed{\frac{5}{4}}$

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