The sum of $10$ terms of the series  $\frac{3}{1^2\times 2^2}+\frac{5}{2^2\times 3^2}+\frac{7}{3^2\times 4^2}+\dots$ is :

Given that

$\frac{3}{1^2\times 2^2}+\frac{5}{2^2\times 3^2}+\frac{7}{3^2\times 4^2}+\dots$

Here $3,5,7,----\in A.P$

First term $a = 3 \&$Common difference $d =2, t_r = a + \left(r-1\right)d = 3 + \left(r-1\right)2 = 2r+1$

$=\sum _{r=1}^{10}\frac{(2r+1)}{r^2·(r+1{)}^2}$

$=\sum _{r=1}^{10}\frac{r^2+2r+1-r^2}{r^2·(r+1{)}^2}$

$=\sum _{r=1}^{10}\frac{(r+1{)}^2-r^2}{r^2·(r+1{)}^2}$

$=\sum _{r=1}^{10}\left\{\frac{1}{r^2}-\frac{1}{(r+1{)}^2}\right\}$

$=\left(\frac{1}{1^2}-\frac{1}{2^2}\right)+\left(\frac{1}{2^2}-\frac{1}{3^2}\right)+\left(\frac{1}{3^2}-\frac{1}{4^2}\right)+\dots ..+\left(\frac{1}{{10}^2}-\frac{1}{{11}^2}\right)=1-\frac{1}{121}=\frac{120}{121}$