I have to prove this inductively, bare with me if the lingo isn't perfect.
Prove that: 1(1!) + 2(2!) + ... + n(n!) = (n+1)! -1
Proof:
Base Case: n=1;
LHS = 1(1!) = 1
RHS = (1+1)! - 1 = 2! - 1 = 1
Done;
Now, assume true for n; Prove true for n=n+1.
Let n= n+1
RHS = ( (n+1)+1)! - 1 = (n + 2)! - 1
LHS = 1(1!) + 2(2!) + ... + n(n!) + (n+1)((n+1)!)
now, since we stated that 1(1!) + 2(2!) + ... + n(n!) = (n+1)! -1 we can sub that in its place such that the equation looks like the following:
LHS = (n+1)! - 1 + (n+1)((n+1)!)
Now here is where I am stuck ... I have tried factoring and combining like terms in every way I can think of ... anyone want to lend a hand =D
TY