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Difference between revisions of "2016 UNCO Math Contest II Problems/Problem 7"
 
Line 5: Line 5:
  
 
== Solution ==
 
== Solution ==
−
First, we perform fractional decomposition on the summed expression.
+
First, we do fraction decomp.
 
Let <cmath>\frac{A}{(n-1)^2}+\frac{B}{(n+1)^2} = \frac{4n}{(n^2-1)^2}</cmath>.
 
Let <cmath>\frac{A}{(n-1)^2}+\frac{B}{(n+1)^2} = \frac{4n}{(n^2-1)^2}</cmath>.
 
Multiplying both sides by <math>(n^2-1)^2</math> and expanding gives <cmath>(A+B)n^2+2(A-B)n+(A+B)=4n</cmath>
 
Multiplying both sides by <math>(n^2-1)^2</math> and expanding gives <cmath>(A+B)n^2+2(A-B)n+(A+B)=4n</cmath>

Latest revision as of 12:16, 9 February 2024

Problem

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


Solution

First, we do fraction decomp. 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 \[\begin{cases} A+B=0\\ A-B=2\end{cases}\]. Adding the two equations gives $2A=2 \implies A=1$, while subtracting the two gives $2B=-2 \implies B=-1$.

Therefore, \[\frac{4n}{(n^2-1)^2}=\frac{1}{(n-1)^2}-\frac{1}{(n+1)^2}\], so \[S =\sum_{n=2}^{\infty} \frac{1}{(n-1)^2}-\frac{1}{(n+1)^2}\] \[= \sum_{n=2}^{\infty} \frac{1}{(n-1)^2} - \sum_{n=2}^{\infty} \frac{1}{(n+1)^2}\]

Writing out the first few terms and rearranging, we have \[\frac{1}{1^2}+\frac{1}{2^2}+\left(\cancel{\frac{1}{3^2}+\frac{1}{4^2}+\cdots } \right)- \left(\cancel{\frac{1}{3^2}+\frac{1}{4^2}+\cdots } \right)\], which telescopes to $\frac{1}{1^2}+\frac{1}{2^2} = \boxed{\frac{5}{4}}$ -NamelyOrange


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 (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
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