The sum of the infinite series $\frac{1}{3 \times 7} + \frac{1}{7 \times 11} + \frac{1}{11 \times 15} + \dots$ is

$\prod\limits_{n = 1}^{10} {\left( {\frac{{6\sum\limits_{i = 0}^n i + 1}}{{6\sum\limits_{j = 0}^n {(j - 1)} + 1}}} \right)} $ is equal to

Let $S_n = \frac{1}{1^3} + \frac{1 + 2}{1^3 + 2^3} + \frac{1 + 2 + 3}{1^3 + 2^3 + 3^3} + \dots + \frac{1 + 2 + \dots + n}{1^3 + 2^3 + \dots + n^3}$. If $100 S_n = n$,then $n$ is equal to:

The sum of $(n + 1)$ terms of $\frac{1}{1} + \frac{1}{1 + 2} + \frac{1}{1 + 2 + 3} + \dots$ is

The sum of $10$ terms of the series $\frac{3}{1^{2} \times 2^{2}}+\frac{5}{2^{2} \times 3^{2}}+\frac{7}{3^{2} \times 4^{2}}+\ldots$ is :

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