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.

Particular case of: http://math.stackexchange.com/q/737402/321264. — StubbornAtom, Aug 26 '16 at 13:29

score 6 · Accepted Answer · edited Aug 26 '16 at 12:02

6

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.

edited Aug 26 '16 at 12:02

b00n heT

answered Aug 26 '16 at 11:30

Olivier Oloa

How did we get ...=[(n+1)-n]n! – geeky me Aug 26 '16 at 11:40
@geekyme Because $(n+1)!=(n+1)\cdot n!$... – 5xum Aug 26 '16 at 11:42
I am still unable to understand because expanding [(n+1)-n]n! would give (n+1)! - nn!. But l.h.s. is (n+1)! - n! – geeky me Aug 26 '16 at 11:48
@geekyme indeed there was a typo in Olivier's answer. See the edit – b00n heT Aug 26 '16 at 12:02
1

Thanks I was also thinking that there was some mistake but I am relatively new so I was not sure . :-) – geeky me Aug 26 '16 at 12:07
1

@b00nheT Thank you very much! – Olivier Oloa Aug 26 '16 at 12:18
@OlivierOloa you are welcome :) – b00n heT Aug 26 '16 at 12:21

score 4 · Answer 2 · answered Aug 26 '16 at 11:32

4

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

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

answered Aug 26 '16 at 11:32

gammatester

2 Answers2