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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
$41600$
$47651$
$41651$
$41671$
Solution
$S_{n}=\frac{n^{2}}{1-\frac{1}{(n+1)^{2}}}=\frac{n(n+1)^{2}}{(n+2)}$
$S_{n}=\frac{n\left(n^{2}+2 n+1\right)}{(n+2)}$
$S_{n}=\frac{n[n(n+2)+1]}{(n+2)}$
$S_{n}=n\left[n+\frac{1}{n+2}\right]$
$S_{n}=n^{2}+\frac{n+2-2}{(n+2)}$
$S_{n}=n^{2}+1-\frac{2}{(n+2)}$
Now $\frac{1}{26}+\sum \limits_{n=1}^{50}\left[\left(n^{2}-n\right)-2\left(\frac{1}{n+2}-\frac{1}{n+1}\right)\right]$
$=\frac{1}{26}+\left[\frac{50 \times 51 \times 101}{6}-\frac{50 \times 51}{2}-2\left(\frac{1}{52}-\frac{1}{2}\right)\right]$
$=41651$
