Let for n = 1, 2, , 50, S _ n be the sum of the infinite geometric...

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

${S}_{n} = \dfrac{{n}^{2}}{1 - \dfrac{1}{{\left(n + 1\right)}^{2}}} = \dfrac{{n}^{2}{\left(n + 1\right)}^{2}}{n\left(n + 2\right)}$

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

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

Now, $\mathop{∑}_{n = 1}^{50}\left({S}_{n} + \dfrac{2}{n + 1} - n - 1\right)$

From equation \left(i\right)$, </p><p>$= \mathop{∑}_{n = 1}^{50}\left({n}^{2} + 1 - \dfrac{2}{n + 2} + \dfrac{2}{n + 1} - n - 1\right)$</p><p>$= \mathop{∑}_{n = 1}^{50}\left({n}^{2} - n\right) + 2\mathop{∑}_{n = 1}^{50}\left(\dfrac{1}{n + 1} - \dfrac{1}{n + 2}\right)$</p><p>$= \mathop{∑}_{n = 1}^{50}{n}^{2} - \mathop{∑}_{n = 1}^{50}n + 2\left[\left(\dfrac{1}{2} - \dfrac{1}{3}\right) + \left(\dfrac{1}{3} - \dfrac{1}{4}\right) + ⋯ + \left(\dfrac{1}{51} - \dfrac{1}{52}\right)\right]$</p><p>$= \dfrac{50\left(51\right)\left(101\right)}{6} - \dfrac{50\left(51\right)}{2} + 2\left[\dfrac{1}{2} - \dfrac{1}{52}\right]$</p><p>$= \left(25\right)\left(17\right)\left(101\right) - 25\left(51\right) + \left(\dfrac{50}{52}\right)$</p><p>$= 25\left(1717 - 51\right) + \dfrac{50}{52} = 25\left(1666\right) + \dfrac{25}{26}$</p><p>$= 41650 + \dfrac{25}{26}$</p><p>$∴ \dfrac{1}{26} + \mathop{∑}_{n = 1}^{50}\left({S}_{n} + \dfrac{2}{n + 1} - n - 1\right) = 41650 + 1 = 41651