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Math Help - Proof by mathematical Induction

  1. #1
    Newbie
    Joined
    Apr 2009
    Posts
    2

    Proof by mathematical Induction

    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?
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  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,681
    Thanks
    613
    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.

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  3. #3
    Newbie
    Joined
    Apr 2009
    Posts
    2
    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)
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