Simplifying both sides of a mathematical induction proof

[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)!

 

[HallsofIvy]

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.

 

[Mono182]

ok, here's 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

 

[ZioX]

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

 

[VietDao29]

(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

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? :)