Induction Method?

Question

Prove that 1(1!) + 2(2!) +.....+n(n!) = (n+1)! -1

Answer 1

True for n=1: 1(1!) = 2! - 1 = 1.

Induction: Given that sum(i=1 to n) i(i!) = (n+1)! - 1, we have

sum(i=1 to n+1) i(i!)
= sum(i=1 to n) i(i!) + (n+1)[(n+1)!]
= (n+1)! -1 + (n+1)[(n+1)!]
= (n+1)[(n+1)!] + (n+1)! - 1 (rearranging terms)
= (n+2)[(n+1)!] - 1
= (n+2)! - 1.

Since the statement is true for n=1 and it being true for n causes it to be true for n+1, it follows by induction that it is true for all n.

Answer 2

First we note that this is trivially true when n=1. Assume that it is true for some n. Then [k=1, n+1]∑kk! = (n+1)(n+1)! + [k=1, n]∑kk! = (n+1)(n+1)! + (n+1)!-1 = (n+1+1)(n+1)! - 1 = (n+2)! - 1, so this formula also holds for n+1. Thus by induction, it holds for all n, and we are done.

Answer 3

Show good for n=2 Ok

Assume true for n=k
1(1!)+2(2!) +...+k(k!) = (k+1)!-1

add both sides (k+1)(k+1)!

1(1!)+2(2!)+...(k+1)(k+1)! = (k+1)!-1+(k+1)(k+1)!
=(k+1)![1+(k+1)]-1
=[(k+1)+1]!-1
qed

Answer 4

Too abstract, can't do sorry. I reckon it's n+2=n