If the sum of the series $\frac{1}{1 \cdot(1+d)} + \frac{1}{(1+d)(1+2d)} + \dots + \frac{1}{(1+9d)(1+10d)}$ is equal to $5$,then $50d$ is equal to:

$\lim _{n \rightarrow \infty} \sum_{r=1}^n \tan ^{-1}\left(\frac{2 r}{r^4+r^2+2}\right) = $

The sum of the series $\frac{1}{x+1}+\frac{2}{x^{2}+1}+\frac{2^{2}}{x^{4}+1}+\ldots+\frac{2^{100}}{x^{2^{100}}+1}$ when $x=2$ is:

Let $S_k, k=1, 2, \ldots, 100$,denote the sum of the infinite geometric series whose first term is $\frac{k-1}{k!}$ and the common ratio is $\frac{1}{k}$. Then the value of $\frac{100^2}{100!} + \sum_{k=1}^{100} |(k^2 - 3k + 1) S_k|$ is

The sum to $50$ terms of the series $\frac{3}{1^2} + \frac{5}{1^2 + 2^2} + \frac{7}{1^2 + 2^2 + 3^2} + \dots$ is:

The sum $\sum\limits_{r = 1}^{10} {({r^2} + 1) \times r!}$ is equal to