Question

Given $S_1=\sum _{n=0}^k\frac{2n+3}{(n+1{)}^2(n+2{)}^2}$ and $S_2=\sum _{n=1}^k\frac{2}{4n^2-1}$

If $S_1-S_2=\frac{363}{37\times 400},$ then $k=$

Answer 1

$S_1=\sum _{n=0}^k\frac{2n+3}{(n+1{)}^2(n+2{)}^2}=\sum _{n=0}^k(\frac{1}{(n+1{)}^2}-\frac{1}{(n+2{)}^2})=(\frac{1}{1}-\frac{1}{4})+(\frac{1}{4}-\frac{1}{9})+(\frac{1}{9}-\frac{1}{16})+\cdots +(\frac{1}{(k+1{)}^2}-\frac{1}{(k+2{)}^2})=1-\frac{1}{(k+2{)}^2}$

$S_2=\sum _{n=1}^k\frac{2}{4n^2-1}=\sum _{n=1}^k(\frac{1}{2n-1}-\frac{1}{2n+1})=(\frac{1}{1}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{5})+(\frac{1}{5}-\frac{1}{7})+\cdots +(\frac{1}{2k-1}-\frac{1}{2k+1})=1-\frac{1}{2k+1}$

So, $S_1-S_2=(1-\frac{1}{(k+2{)}^2})-(1-\frac{1}{2k+1})\Rightarrow \frac{(k+1{)}^2+2}{(2k+1)(k+2{)}^2}=\frac{363}{37\times 400}\Rightarrow (k+2{)}^2=400\Rightarrow k=18$