What is the sum of the first $n$ terms of the series whose $k$-th term is $k! \times k$?

Find the sum to $n$ terms of the series $\frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} + \ldots$

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:

For any $n \in N$,$\frac{1}{2 \cdot 5} + \frac{1}{5 \cdot 8} + \ldots + \frac{1}{(3n-1)(3n+2)} = $

If ${a_k} = \frac{1}{{k(k + 1)}}$ for $k = 1, 2, 3, 4, ..., n$,then ${\left( {\sum\limits_{k = 1}^n {{a_k}} } \right)^2} = $

If the sum of the series $\frac{1}{1 \cdot(1+d)} + \frac{1}{(1+d)(1+2d)} + \dots + \frac{1}{(1+9d)(1+10d)}$ is equal to $5$,then $50d$ is equal to: