Question

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

Answer 1

$\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+......nterms=\frac{1+2}{1^2.2^2}+\frac{2+3}{2^2.3^2}+\frac{3+4}{3^2.4^2}+.......+\frac{n(n+1)}{n^2{(n+1)}^2}=(\frac{1}{1^2}-\frac{1}{2^2})+(\frac{1}{2^2}-\frac{1}{3^2})+(\frac{1}{3^2}-\frac{1}{4^2})+.........+(\frac{1}{n^2}-\frac{1}{{(n+1)}^2})=\frac{1}{1^2}-\frac{1}{{(n+1)}^2}=\frac{n^2+2n}{{(n+1)}^2}$