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. | ||
+ | |||
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> | + | 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>. |
+ | 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{4n}{(n^2-1)^2}</math> | ||
<math>\boxed{\frac{5}{4}}</math> | <math>\boxed{\frac{5}{4}}</math> |
Revision as of 18:27, 21 October 2023
Contents
[hide]Problem
Evaluate
Solution
First, we perform fractional decomposition on the summed expression.
Let . 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
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 |