$\frac{1}{2 \cdot 5} + \frac{1}{5 \cdot 8} + \frac{1}{8 \cdot 11} + \ldots + \frac{1}{(3n-1)(3n+2)}$ is equal to

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

$\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

The value of the infinite product $\prod\limits_{n = 2}^\infty {\left( {1 - \frac{1}{{{n^2}}}} \right)}$ is

If $\frac{1}{(20-a)(40-a)}+\frac{1}{(40-a)(60-a)}+\ldots+\frac{1}{(180-a)(200-a)}=\frac{1}{256}$,then the maximum value of $a$ is.

The sum of the infinite series $\cot ^{-1}\left(\frac{7}{4}\right)+\cot ^{-1}\left(\frac{19}{4}\right)+\cot ^{-1}\left(\frac{39}{4}\right)+\cot ^{-1}\left(\frac{67}{4}\right)+\ldots \ldots$ is :-