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

Assertion $(A)$: $1+(1+2+4)+(4+6+9)+(9+12+16)+\ldots+(81+90+100)=1000$
Reason $(R)$: $\sum_{r=1}^n(r^3-(r-1)^3)=n^3$ for any natural number $n$.

The sum of the series $\frac{1}{\sqrt{1} + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + ... + \frac{1}{\sqrt{n^2 - 1} + \sqrt{n^2}}$ equals

Find the sum of the series: $1 \cdot 1! + 2 \cdot 2! + 3 \cdot 3! + \dots + n \cdot n!$

If $\sum_{r=1}^{n} T_{r} = \frac{(2n-1)(2n+1)(2n+3)(2n+5)}{64}$,then $\lim_{n \rightarrow \infty} \sum_{r=1}^{n} \left(\frac{1}{T_{r}}\right)$ is equal to :

$\frac{1}{1 \times 5} + \frac{1}{5 \times 9} + \frac{1}{9 \times 13} + \ldots$ up to $n$ terms $=$