Evaluate $\sum_{n=1}^{50}n\cdot n!$.

Question

Evaluate $\displaystyle\sum_{n=1}^{50}n \!\cdot\! n!$

I tried to write the sum like: $1+2\!\cdot\!2!+3\!\cdot\!3!+4\!\cdot\!4!+5\!\cdot\!5!+\ldots=1+4+18+96+600\ldots$

I can find a recursion like $t_{n+1}=t_n(n+2)+n!$ where $t_n=n\!\cdot\!n!$, so:

$1\!\cdot\!3+1=4$

$4\!\cdot\!4+(1\!\cdot\!2)=18$

$18\!\cdot\!5+(1\!\cdot\!2\!\cdot\!3)=96$

$96\!\cdot\!6+(1\!\cdot\!2\!\cdot\!3\!\cdot\!4)=600$

But I'm not sure how to proceed further or if this is the way to do it. Any help is appreciated.

Answer 1

The trick is as follows: $$\sum_{n=1}^{50} n!n=\sum_{n=1}^{50} n!(n+1-1)=\sum_{n=1}^{50} [(n+1)!-n!]=51!-1$$

Answer 2

Hint:$n\cdot n! = (n+1)n! - n! = (n+1)! - n! $

Telescoping sum.