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put this solution on YOUR website! 1(1!)+2(2!)+3(3!)+...+n(n!) = (n+1)!-1
First we prove it's true for n=1
1(1!) = 1(1) = 1 and (1+1)!-1 = 2!-1 = 2-1 = 1
Now we assume it's true for n=k
(1) 1(1!)+2(2!)+3(3!)+...+k(k!) = (k+1)!-1
We need to show that
(2) 1(1!)+2(2!)+3(3!)+...+(k+1)(k+1)! ≟ (k+2)!-1
We add (k+1)(k+1)! to both sides of (1)
(1) 1(1!)+2(2!)+3(3!)+...+k(k!)+(k+1)(k+1)! = (k+1)!-1+(k+1)(k+1)! =
= (k+1)!+(k+1)(k+1)!-1 =
= (k+1)![1+(k+1)]-1 =
= (k+1)![1+k+1]-1 =
= (k+1)!(k+2)-1 =
= (k+2)!-1
So the truth of (1) implies the truth of (2). So the induction is complete.
Edwin