# Mathematical induction question

1. Jun 3, 2007

### Mono182

i'm on the last part of this question involving mathematical induction and i can't get the left side to equal the right saide. can anyone help me out?

right side: [(k+1)+1]! - 1

left side: (k+1)! - 1 + (k+1) + (k+1)!

2. Jun 3, 2007

### HallsofIvy

Staff Emeritus
What you mean you "can't get the left side to equal the right side"? Show what equations you do have so we can, if necessary, point out an error.

3. Jun 3, 2007

### Mono182

ok, heres the full question:

Using mathematical induction, prove that the following statements are true for n > 1, n = 1.

Q: 1 X 1! + 2 X 2! + ... + n X n! = (n + 1)! - 1

A: step 1: solve for n = 1

1 X 1! = (1 + 1)! - 1
1 = 1

step 2: assume true for n = k

therefore 1 X 1! + 2 X 2! + ... + k X k! = (k + 1)! - 1

step 3: sove for n = k + 1

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

since 1 X 1! + 2 X 2! + ... + k X k! = (k + 1)! - 1 , then

(k + 1)! - 1 + (k+1) X (k + 1)! = [(k+1) + 1)] - 1

now what i'm having trouble with is proving that the last statement is true

4. Jun 4, 2007

### ZioX

Factor out (k+1)! (ignoring the -1)

5. Jun 4, 2007

### VietDao29

Yup, so far so good.
As Ziox's pointed out, you should factor out (k + 1)!
The LHS, and the RHS, both have "-1", so they cancel each other out, leaving you with:
(k + 1)! + (k + 1) (k + 1)! = (k + 2)!

You should also notice that:
n! = n (n - 1) (n - 2) ... 2 . 1 = n (n - 1)!

Can you go from here? :)