Ticklish series

$n(n!) = ((n+1)-1)n! = (n+1)!-n!$ . Hence $\sum_{n=1}^{50}n(n!) = (51!-50!)+(50!-49!)+\dots+(2!-1!) = 51!-1$

1×1!=(2-1)×1!=2!-1!

2×2!=(3-2)×2!=3!-2!

3×3!=.............=4!-3!

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49×49=50!-49!

50×50=51!-50!

On adding... we get 50!-1

I will prove, using mathematical induction, $\sum_{n=1}^kn\cdot n!=(k+1)!-1.$

Base case ( $P(k=1)$ ): $\sum_{n=1}^1 n\cdot n! = 1\cdot 1! = 1 = (1+1)!-1$

Induction step: Assume $P(k)$ . That is, assume $\sum_{n=1}^kn\cdot n!=(k+1)!-1$ for some $k\geq 1.$

$\Rightarrow \sum_{n=1}^{k+1} n\cdot n!= (k+1)!-1+(k+1)(k+1)! = (k+2)(k+1)!-1=(k+2)!-1 \Leftrightarrow P(k+1)$

Thus, $\sum_{n=1}^kn\cdot n!=(k+1)!-1$ for all $k\geq 1.$