Proof by mathematical Induction

**dan7** · April 20th 2009, 06:08 AM

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?

**Soroban** · April 20th 2009, 07:19 AM

Hello, dan7!

I'll show you my approach to Induction.

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

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.

**dan7** · April 20th 2009, 07:33 AM

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)

Thread: Proof by mathematical Induction

LinkBack

Thread Tools

Display

Proof by mathematical Induction

Similar Math Help Forum Discussions

Proof by Mathematical Induction

Proof by mathematical induction.

Mathematical Induction Proof

Mathematical Induction Proof

proof and mathematical induction

Search Tags