Difference between revisions of "2016 UNCO Math Contest II Problems/Problem 7"
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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> | ||
+ | |||
== Solution == | == Solution == | ||
− | <math>\frac{5}{4}</math> | + | 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>. | ||
+ | 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\ | ||
+ | A-B=2\end{cases}</math>, which has the solution <math>(A,B)=(1,-1)</math>. | ||
+ | |||
+ | <math>\boxed{\frac{5}{4}}</math> | ||
==Solution 2== | ==Solution 2== |
Revision as of 18:06, 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 , which has the solution .
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 |