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 $\sum_{r=1}^{25} \left( \frac{r}{r^{4}+r^{2}+1} \right) = \frac{p}{q}$ where $p$ and $q$ are positive integers such that $\gcd(p,q)=1$, then $p+q$ is equal to . . . . . . .

The value of $\sum\limits_{n = 2}^\infty {\frac{n}{{1 + {n^2}\left( {{n^2} - 2} \right)}}} $ is equal to

The sum of $n$ terms of the series $\frac{1}{1 + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{5}} + \frac{1}{\sqrt{5} + \sqrt{7}} + \dots$ is

If $\frac{1}{1 \cdot 5}+\frac{1}{5 \cdot 9}+\frac{1}{9 \cdot 13}+\ldots$ to $n$ terms $= \frac{27}{109}$,then $n = $

If $t_{n} = \frac{1}{4}(n+2)(n+3)$,$n \in N$,then which one of the following is true?
Assertion $(A)$ : $\frac{1}{t_1} + \frac{1}{t_2} + \ldots + \frac{1}{t_{2003}} = \frac{2003}{3009}$
Reason $(R)$ : $\frac{1}{t_1} + \frac{1}{t_2} + \ldots + \frac{1}{t_{n}} = \frac{4n}{3(n+3)}$