# Thread: Proof by mathematical Induction

1. ## Proof by mathematical Induction

Hello, I need some help with a proof by mathematical induction. Any help as soon as possible is appreciated

The equation is:

n
Σ k!.k = (n+1)! - 1
k=1

Now, here is what I have so far
n+1
Σ k! * k
k=1

goes to
n
Σ k! * k + (n+1)! * (n+1)
k=1

As for the right hand side, it becomes
((n+1)+1)! - 1
=(n+2)! - 1

Now this is the point where i am stuck, how does
n
Σ k! * k + (n+1)! * (n+1)
k=1
become (n+2) -1 ?

or have I done something else wrong?

2. Hello, dan7!

I'll show you my approach to Induction.

Prove by mathematical induction: . $
\sum^n_{k=1} k(k!) \;=\;(n+1)!-1$

Verify $S(1)\!:\;\;1(1!) \:=\:2! - 1 \:=\:1$ . . . true!

Assume $S(k)\!:\;\;1(1!) + 2(2!) + 3(3!) + \hdots + k(k!) \;=\;(k+1)! - 1$

Add $(k+1)(k+1)!$ to both sides:

. . $1(1!) \;+\; 2(2!) \;+\; 3(3!) \;+\; \hdots \;+\; k(k!) \;+\; {\color{blue}(k+1)(k+1)!} \;=\;(k+1)! \;- \;1 \;+\; {\color{blue}(k+1)(k+1)!}$

The right side is: . $(k+1)! + (k+1)(k+1)! - 1$

. . . . . . . Factor: . $(k+1)!\,\bigg[1 + (k+1)\bigg] - 1 \;=\;(k+1)!(k+2) - 1 \;=\;(k+2)!-1$

We have shown that $S(k+1)$ is true:

. . $1(1!) + 2(2!) + 3(3!) + \hdots + (k+1)(k+1)! \;=\;(k+2)! - 1$

The inductive proof is complete.

3. Thank you VERY much, I see now where i was going wrong (I was trying to factor the k(k!) instead of the right side)