Let for $n=1,2,\dots \dots ,50,{ S}_n$ be the sum of the infinite geometric progression whose first term is $n^2$ and whose common ratio is $\frac{1}{{\left(n+1\right)}^2}$. Then the value of $\frac{1}{26}+\sum _{n=1}^{50}\left(S_n+\frac{2}{n+1}-n-1\right)$ is equal to

We have, $a=n^2, r=\frac{1}{{\left(n+1\right)}^2}$

$S_n=\frac{n^2}{1-\frac{1}{{\left(n+1\right)}^2}}=\frac{n^2{\left(n+1\right)}^2}{n\left(n+2\right)}$

$=\frac{n\left[n\left(n+2\right)+1\right]}{n+2}=n^2+\frac{n}{n+2}$

$=n^2+\frac{n+2-2}{n+2}$  $=n^2+1-\frac{2}{n+2} ...\left(\mathrm{i}\right)$

Now, $\sum _{n=1}^{50}\left(S_n+\frac{2}{n+1}-n-1\right)$

From equation $\left(i\right)$, 

$=\sum _{n=1}^{50}\left(n^2+1-\frac{2}{n+2}+\frac{2}{n+1}-n-1\right)$

$=\sum _{n=1}^{50}\left(n^2-n\right)+2\sum _{n=1}^{50}\left(\frac{1}{n+1}-\frac{1}{n+2}\right)$

$=\sum _{n=1}^{50}{n}^2-\sum _{n=1}^{50}n+2\left[\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\cdots +\left(\frac{1}{51}-\frac{1}{52}\right)\right]$

$=\frac{50\left(51\right)\left(101\right)}{6}-\frac{50\left(51\right)}{2}+2\left[\frac{1}{2}-\frac{1}{52}\right]$

$=\left(25\right)\left(17\right)\left(101\right)-25\left(51\right)+\left(\frac{50}{52}\right)$

$=25\left(1717-51\right)+\frac{50}{52}$  $=25\left(1666\right)+\frac{25}{26}$

$=41650+\frac{25}{26}$

$\therefore \frac{1}{26}+\sum _{n=1}^{50}\left(S_n+\frac{2}{n+1}-n-1\right)=41650+1=41651$