The value of $\lim _{n \rightarrow \infty} \left( \sum_{k=1}^n \frac{k^3+6 k^2+11 k+5}{(k+3)!} \right)$ is:

$1 + \sum\limits_{r = 0}^{22} {\left\{ {r\left( {r + 2} \right) + 1} \right\}} \cdot r! = k!$,then the number of divisors of $k$ is

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 = $

If $\frac{1}{2 \times 4} + \frac{1}{4 \times 6} + \frac{1}{6 \times 8} + \dots (n \text{ terms}) = \frac{k n}{n+1}$,then $k$ is equal to

If ${x_1}, {x_2}, {x_3}, \dots, {x_n}$ are in $A.P.$ whose common difference is $\alpha$,then the value of $\sin \alpha (\sec {x_1} \sec {x_2} + \sec {x_2} \sec {x_3} + \dots + \sec {x_{n-1}} \sec {x_n}) = $

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