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Mathematical induction question

  1. Jun 3, 2007 #1
    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. jcsd
  3. Jun 3, 2007 #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.
     
  4. Jun 3, 2007 #3
    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
     
  5. Jun 4, 2007 #4
    Factor out (k+1)! (ignoring the -1)
     
  6. Jun 4, 2007 #5

    VietDao29

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    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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