Let for n = 1, 2, ldots, 50,S_{n} be the sum | vedclass

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(C) The sum of an infinite geometric progression is $S = \frac{a}{1-r}$.Given $a = n^{2}$ and $r = \frac{1}{(n+1)^{2}}$,we have $S_{n} = \frac{n^{2}}{

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$41651$

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(C) The sum of an infinite geometric progression is $S = \frac{a}{1-r}$.Given $a = n^{2}$ and $r = \frac{1}{(n+1)^{2}}$,we have $S_{n} = \frac{n^{2}}{1 - \frac{1}{(n+1)^{2}}} = \frac{n^{2}(n+1)^{2}}{(n+1)^{2}-1} = \frac{n^{2}(n+1)^{2}}{n^{2}+2n} = \frac{n(n+1)^{2}}{n+2}$.We can rewrite $S_{n}$ as $S_{n} = \frac{n(n^{2}+2n+1)}{n+2} = \frac{n(n(n+2)+1)}{n+2} = n^{2} + \frac{n}{n+2} = n^{2} + \frac{n+2-2}{n+2} = n^{2} + 1 - \frac{2}{n+2}$.Now,substitute this into the summation: $\sum_{n=1}^{50} (S_{n} + \frac{2}{n+1} - n - 1) = \sum_{n=1}^{50} (n^{2} + 1 - \frac{2}{n+2} + \frac{2}{n+1} - n - 1) = \sum_{n=1}^{50} (n^{2} - n + 2(\frac{1}{n+1} - \frac{1}{n+2}))$.This is a telescoping sum: $\sum_{n=1}^{50} (n^{2} - n) + 2 \sum_{n=1}^{50} (\frac{1}{n+1} - \frac{1}{n+2})$.$= (\frac{50 	imes 51 	imes 101}{6} - \frac{50 	imes 51}{2}) + 2(\frac{1}{2} - \frac{1}{52}) = (42925 - 1275) + 2(\frac{25}{52}) = 41650 + \frac{25}{26}$.Adding the term $\frac{1}{26}$ from the original expression: $41650 + \frac{25}{26} + \frac{1}{26} = 41650 + 1 = 41651$.

Let for $n = 1, 2, \ldots, 50$,$S_{n}$ 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

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