Question

Sum: $\frac{3}{1^2{.2}^2}+\frac{5}{2^2{.3}^2}+\frac{7}{3^2{.4}^2}+.......$ to n terms.

Answer 1

$\left(\frac{3}{1^2\cdot 2^2}\right)+\left(\frac{5}{2^2\cdot 3^2}\right)+\left(\frac{7}{3^2\cdot 4^2}\right)+.........\left(\frac{2n+1}{n^2(n+1{)}^2}\right)now\left(\frac{3}{1^2\cdot 2^2}\right)=\left(\frac{2^2-1^2}{1^2\cdot 2^2}\right)=\left(\frac{1}{1^2}\right)-\left(\frac{1}{2^2}\right)\left(\frac{5}{2^2\cdot 3^2}\right)=\left(\frac{3^2-2^2}{2^2\cdot 3^2}\right)=\left(\frac{1}{1^2}\right)-\left(\frac{1}{3^2}\right)\therefore sum=\left(\right(\frac{1}{1^2})-(\frac{1}{2^2})+(\frac{1}{2^2})-(\frac{1}{3^2}\left)\right)+.......+\left[\right(\frac{1}{n^2})-(\frac{1}{(n+1{)}^2}\left)\right]=1-\left(\frac{1}{(n+1{)}^2}\right)=\left(\frac{n^2+2n}{(n+1{)}^2}\right)$