Question

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 $\frac{1}{(n+1{)}^2}$. Then the value of $\frac{1}{26}+\sum _{n=1}^{50}\left(S_n+\frac{2}{n+1}-n-1\right)$ is equal to

Answer 1

Answer: 41651

$S_n=\frac{n^2}{1-\frac{1}{(n+1{)}^2}}=\frac{n(n+1{)}^2}{(n+2)}$
$S_n=\frac{n\left(n^2+2n+1\right)}{(n+2)}$
$S_n=\frac{n[n(n+2)+1]}{(n+2)}$
$S_n=n\left[n+\frac{1}{n+2}\right]$
$S_n=n^2+\frac{n+2-2}{(n+2)}$
$S_n=n^2+1-\frac{2}{(n+2)}$
Now $\frac{1}{26}+\sum _{n=1}^{50}\left[\left(n^2-n\right)-2\left(\frac{1}{n+2}-\frac{1}{n+1}\right)\right]$
$=\frac{1}{26}+\left[\frac{50\times 51\times 101}{6}-\frac{50\times 51}{2}-2\left(\frac{1}{52}-\frac{1}{2}\right)\right]$
$=41651$