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Thread: 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
    12,028
    Thanks
    848
    Hello, dan7!

    I'll show you my approach to Induction.


    Prove by mathematical induction: .$\displaystyle
    \sum^n_{k=1} k(k!) \;=\;(n+1)!-1$

    Verify $\displaystyle S(1)\!:\;\;1(1!) \:=\:2! - 1 \:=\:1$ . . . true!


    Assume $\displaystyle S(k)\!:\;\;1(1!) + 2(2!) + 3(3!) + \hdots + k(k!) \;=\;(k+1)! - 1$


    Add $\displaystyle (k+1)(k+1)!$ to both sides:

    . . $\displaystyle 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: .$\displaystyle (k+1)! + (k+1)(k+1)! - 1$

    . . . . . . . Factor: .$\displaystyle (k+1)!\,\bigg[1 + (k+1)\bigg] - 1 \;=\;(k+1)!(k+2) - 1 \;=\;(k+2)!-1$



    We have shown that $\displaystyle S(k+1)$ is true:

    . . $\displaystyle 1(1!) + 2(2!) + 3(3!) + \hdots + (k+1)(k+1)! \;=\;(k+2)! - 1$

    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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