Let for n =1,2, ldots ldots, 50, S _{ a } be | vedclass

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Question

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

Vedclass · Accepted Answer

$41651$

Vedclass · Answer

$S_{n}=\frac{n^{2}}{1-\frac{1}{(n+1)^{2}}}=\frac{n(n+1)^{2}}{(n+2)}$

$S_{n}=\frac{n\left(n^{2}+2 n+1ight)}{(n+2)}$

$S_{n}=\frac{n[n(n+2)+1]}{(n+2)}$

$S_{n}=n\left[n+\frac{1}{n+2}ight]$

$S_{n}=n^{2}+\frac{n+2-2}{(n+2)}$

$S_{n}=n^{2}+1-\frac{2}{(n+2)}$

Now $\frac{1}{26}+\sum \limits_{n=1}^{50}\left[\left(n^{2}-night)-2\left(\frac{1}{n+2}-\frac{1}{n+1}ight)ight]$

$=\frac{1}{26}+\left[\frac{50 	imes 51 	imes 101}{6}-\frac{50 	imes 51}{2}-2\left(\frac{1}{52}-\frac{1}{2}ight)ight]$

$=41651$

Let for $n =1,2, \ldots \ldots, 50, S _{ a }$ 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 $\frac{1}{26}+\sum\limits_{n=1}^{50}\left(S_{n}+\frac{2}{n+1}-n-1\right)$ is equal to

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