Question

Let for n = 1, 2, …, 50, Sn be the sum of the infinite geometric progression whose first term is n2 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

The corrrect answer is 41651
$S_n=\frac{n2}{1−\frac{1}{(n+1)^2}}=\frac{n(n+1)^2}{n+2}=(n2+1)−\frac{2}{n+2}$
Now
$\frac{1}{26} + \sum_{n=1}^{50} \left(S_n+\frac{2}{n+1}-n-1 \right)$
$\frac{1}{26} + \sum_{n=1}^{50} \left( n^2 - n + 2 \left( \frac{1}{n+1} - \frac{1}{n+2} \right) \right)$
$\frac{1}{26} + \frac{50 \times 51 \times 101}{6} - \frac{50 \times 51}{2} + 2 \left( \frac{1}{2} - \frac{1}{52} \right)$
= 1 + 25 × 17 (101 – 3)
= 41651