Mathematical induction question

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  • Thread starter Mono182
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  • #1
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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)!
 

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  • #2
HallsofIvy
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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
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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
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Factor out (k+1)! (ignoring the -1)
 
  • #5
VietDao29
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(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. :smile:
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? :)
 

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