For n=1,2,,50, let S_n be the sum of the infinite geometric progression whose first term i…

Question

For $n=1,2,\ldots,50$, let $S_n$ be the sum of the infinite geometric progression whose first term is $n^2$ and common ratio is $\dfrac{1}{(n+1)^2}$. 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: 41651

Accepted Answer

$$S_n = \frac{n^2}{1-\frac{1}{(n+1)^2}} = \frac{n^2(n+1)^2}{n(n+2)} = \frac{n(n+1)^2}{n+2} = (n^2+1)-\frac{2}{n+2}.$$

So $S_n + \dfrac{2}{n+1}-n-1 = (n^2-n) + 2\left(\dfrac{1}{n+1}-\dfrac{1}{n+2}\right)$, and summing (with the telescoping part):

$$\frac{1}{26}+\sum_{n=1}^{50}(n^2-n) + 2\left(\frac12-\frac{1}{52}\right) = \frac{50\times51\times101}{6}-\frac{50\times51}{2}+1 = 41651.$$

Solutions · 1