Difference between revisions of "2016 UNCO Math Contest II Problems/Problem 7"
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== Solution == | == Solution == | ||
First, we perform fractional decomposition on the summed expression. | 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>. | 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 <math>(A+B)n^2+2(A-B)n+(A+B)=4n</math> | 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> | ||
Therefore, we have the system of equations <math>\begin{cases} A+B=0\\ | Therefore, we have the system of equations <math>\begin{cases} A+B=0\\ | ||
A-B=2\end{cases}</math>. Adding the two equations gives <math>2A=2 \implies A=1</math>, while subtracting the two gives <math>2B=-2 \implies B=-1</math>. | A-B=2\end{cases}</math>. Adding the two equations gives <math>2A=2 \implies A=1</math>, while subtracting the two gives <math>2B=-2 \implies B=-1</math>. | ||
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| − | <math>\boxed{\frac{5}{4}}</math> | + | Therefore, <math>\frac{4n}{(n^2-1)^2}=\frac{1}{(n-1)^2}-\frac{1}{(n+1)^2}</math>, so <math>S =\sum_{n=2}^{\infty} \frac{1}{(n-1)^2}-\frac{1}{(n+1)^2}</math> <math>= \sum_{n=2}^{\infty} \frac{1}{(n-1)^2} - \sum_{n=2}^{\infty} \frac{1}{(n+1)^2}</math> |
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| + | Writing out the first few terms, we have | ||
| + | <math>\left( \frac{1}{1^2}+\frac{1}{2^2}\cancel{+\frac{1}{3^2}+\frac{1}{4^2}+\cdots } \right)- \left\cancel{+\frac{1}{3^2}+\frac{1}{4^2}+\cdots } \right) = \frac{1}{1^2}+\frac{1}{2^2} = \boxed{\frac{5}{4}}</math> | ||
==Solution 2== | ==Solution 2== | ||
Revision as of 11:56, 23 October 2023
Contents
Problem
Evaluate
![\[S =\sum_{n=2}^{\infty} \frac{4n}{(n^2-1)^2}\]](http://latex.artofproblemsolving.com/5/8/f/58f783b714712db7d5b19d0f2b8d4de5d847cb83.png)
Solution
First, we perform fractional decomposition on the summed expression.
Let ![\[\frac{A}{(n-1)^2}+\frac{B}{(n+1)^2} = \frac{4n}{(n^2-1)^2}\]](http://latex.artofproblemsolving.com/6/e/c/6ecc21bc45e4b4c7564eee920e3f10cffb2c529d.png)
.
Multiplying both sides by 
and expanding gives 
Therefore, we have the system of equations 
. Adding the two equations gives 
, while subtracting the two gives 
.
Therefore, 
, so 
Writing out the first few terms, we have $\left( \frac{1}{1^2}+\frac{1}{2^2}\cancel{+\frac{1}{3^2}+\frac{1}{4^2}+\cdots } \right)- \left\cancel{+\frac{1}{3^2}+\frac{1}{4^2}+\cdots } \right) = \frac{1}{1^2}+\frac{1}{2^2} = \boxed{\frac{5}{4}}$ (Error compiling LaTeX. Unknown error_msg)
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 | ||
