Difference between revisions of "2016 UNCO Math Contest II Problems/Problem 7"
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== Solution == | == Solution == | ||
| − | First, we | + | 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 < | + | 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> |
| − | Therefore, we have the system of equations < | + | Therefore, we have the system of equations <cmath>\begin{cases} A+B=0\\ |
| − | A-B=2\end{cases}</math>, | + | A-B=2\end{cases}</cmath>. 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, <cmath>\frac{4n}{(n^2-1)^2}=\frac{1}{(n-1)^2}-\frac{1}{(n+1)^2}</cmath>, so <cmath>S =\sum_{n=2}^{\infty} \frac{1}{(n-1)^2}-\frac{1}{(n+1)^2}</cmath> <cmath>= \sum_{n=2}^{\infty} \frac{1}{(n-1)^2} - \sum_{n=2}^{\infty} \frac{1}{(n+1)^2}</cmath> | ||
| + | |||
| + | Writing out the first few terms and rearranging, we have | ||
| + | <cmath>\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)</cmath>, which telescopes to <math>\frac{1}{1^2}+\frac{1}{2^2} = \boxed{\frac{5}{4}}</math> | ||
| + | -NamelyOrange | ||
| − | |||
==Solution 2== | ==Solution 2== | ||
Latest revision as of 12:16, 9 February 2024
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 do fraction decomp.
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 ![\[(A+B)n^2+2(A-B)n+(A+B)=4n\]](http://latex.artofproblemsolving.com/1/d/3/1d3b7f55361619c1282747e85ada5c15e1dc1f21.png)
Therefore, we have the system of equations ![\[\begin{cases} A+B=0\\ A-B=2\end{cases}\]](http://latex.artofproblemsolving.com/8/2/3/82309de3359fbf399f668755b29a65cb5af35304.png)
. Adding the two equations gives 
, while subtracting the two gives 
.
Therefore, ![\[\frac{4n}{(n^2-1)^2}=\frac{1}{(n-1)^2}-\frac{1}{(n+1)^2}\]](http://latex.artofproblemsolving.com/a/6/6/a66bb9d6e21f94deec67c24d954a50a3292a65bd.png)
, so ![\[S =\sum_{n=2}^{\infty} \frac{1}{(n-1)^2}-\frac{1}{(n+1)^2}\]](http://latex.artofproblemsolving.com/1/b/9/1b96cca5107e46e04fc4bfd59c32f000e76eb39a.png)
![\[= \sum_{n=2}^{\infty} \frac{1}{(n-1)^2} - \sum_{n=2}^{\infty} \frac{1}{(n+1)^2}\]](http://latex.artofproblemsolving.com/b/b/b/bbbc76e471d2093500f1fecf46dcc2365b00aad9.png)
Writing out the first few terms and rearranging, we have
![\[\frac{1}{1^2}+\frac{1}{2^2}+\left(\cancel{\frac{1}{3^2}+\frac{1}{4^2}+\cdots } \right)- \left(\cancel{\frac{1}{3^2}+\frac{1}{4^2}+\cdots } \right)\]](http://latex.artofproblemsolving.com/5/9/f/59fc88399fcd30fe7d95ef601156fc4324ea19b1.png)
, which telescopes to 
-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 | ||
