PLEASE HELP 30 POINTS 1(1!)+2(2!)+3(3!)+...100(100!)=?

Question

asked Sep 26, 2020 91.0k views

2 Answers

Justin Grant · Answer 1 · 2020-10-03T11:05:33+0000

A more general result:

$1\cdot1!+2\cdot2!+\cdots+(n-1)\cdot(n-1)!+n\cdot n!=(n+1)!-1$

Why is this true?

On the left side,

$1\cdot1!=(2-1)\cdot1!=2\cdot1!-1\cdot1!=2!-1!$

$2\cdot2!=(3-1)\cdot2!=3\cdot2!-1\cdot2!=3!-2!$

$3\cdot3!=(4-1)\cdot3!=4\cdot3!-1\cdot3!=4!-3!$

and so on. Adding will lead to cancellation of all terms in the middle, leaving you with
(n+1)!-1 .

Then when
n=100 , the value of the sum is
101!-1 .