Find the sum of $ (1 \times 1!) + (2 \times 2!) + (3 \times 3!) + \cdots+ (50\times50!)$

Question

What will be the sum of following series

$$ (1 \times 1!) + (2 \times 2!) + (3 \times 3!) + \cdots+ (50\times50!)$$

Is there any general solution for $n$ terms?

I have tried writing nth term and then summing up but I was unable to write the nth term only. So now I have no idea how to do this question.

Answer 1

Hint. One may observe that $$ (n+1)!-n!=[(n+1)-1]\cdot n!=n \cdot n!, \quad n=0,1,2,\cdots, $$ getting a telescoping sum here.

Answer 2

You have $$\sum_{k=1}^n kk! = (n+1)!-1$$ Proof via induction, with the induction step:

$$\sum_{k=1}^{n+1} kk! = \sum_{k=1}^n kk! + (n+1)(n+1)! = (n+1)!-1 + (n+1)(n+1)! =(1+n+1)(n+1)! - 1 = (n+2)!-1$$