The sum $\sum_{n=1}^{21} \frac{3}{(4n-1)(4n+3)}$ is equal to

The sum $1 \times 1! + 2 \times 2! + \ldots + 50 \times 50!$ equals

Find the sum of the series $1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + 3 \cdot 4 \cdot 5 + \dots$ up to $n$ terms.

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:

$\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

Let $S_n = \frac{1}{1^3} + \frac{1 + 2}{1^3 + 2^3} + \frac{1 + 2 + 3}{1^3 + 2^3 + 3^3} + \dots + \frac{1 + 2 + \dots + n}{1^3 + 2^3 + \dots + n^3}$. If $100 S_n = n$,then $n$ is equal to: