Question

Find the sum of the series ${\sum }_{r=1}^{n}r\times r!$

Answer 1

We have,

$\begin{array}{c}{\sum }_{r=1}^{n}r\times r!={\sum }_{r=1}^n\left(r+1-1\right)r!\\ {\sum }_{r=1}^n\left[\left(r+1\right)!-r!\right]\\ =\left(2!-1!\right)+\left(3!-2!\right)+.......+\left[(n+1)!-n!\right]\\ =\left(n+1\right)!-1!\\ =\left(n+1\right)!-1\end{array}$

Hence, this is the answer.