If ${T_n} = ({n^2} + 1)n!$ and ${S_n} = {T_1} + {T_2} + {T_3} + ...... + {T_n}$. Let $\frac{{{T_{10}}}}{{{S_{10}}}} = \frac{a}{b}$ where $a$ and $b$ are relatively prime natural numbers,then the value of $(b - a)$ is

If the sum of the first $10$ terms of the series $\frac{4(1)}{1+4(1)^4}+\frac{4(2)}{1+4(2)^4}+\frac{4(3)}{1+4(3)^4}+\ldots$ is $\frac{m}{n}$,where $\operatorname{gcd}(m, n)=1$,then $m+n$ is equal to . . . . . . .

Let the greatest common divisor of $m$ and $n$ be $1$. If $\frac{1}{1 \cdot 7} + \frac{1}{7 \cdot 13} + \frac{1}{13 \cdot 19} + \dots$ up to $20$ terms $= \frac{m}{n}$,then $5m + 2n = $

The value of $\sum\limits_{r = 1}^\infty {{{\tan }^{ - 1}}\left( {\frac{3}{{{r^2} - r + 9}}} \right)} $ is-

The sum of the series $1 + \frac{1}{1 + 2} + \frac{1}{1 + 2 + 3} + \dots$ up to $10$ terms is:

If $S$ is the sum of the first $10$ terms of the series $\tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{1}{13}\right)+\tan ^{-1}\left(\frac{1}{21}\right)+\ldots$ then $\tan ( S )$ is equal to