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Math Help - Induction Proof

  1. #1
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    Induction Proof

    Hi all,

    Prove by induction that \sum_{r=1}^n r(r!)= (n+1)!-1

    for all integers n>=1
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  2. #2
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    Inductive base case.

    Let n=1. Then you have for LHS
    \sum_{r=1}^1 r(r!) = 1(1!) = 1

    And RHS
    (1+1)! - 1 = 2! - 1 = 2-1 = 1. So hold for n=1.

    Assume it holds for n, i.e \sum_{r=1}^n r(r!) = (n+1)! - 1

    Try to prove it works for n+1

    So for n+1 we want to get \sum_{r=1}^{n+1} r(r!) = ((n+1)+1)! - 1 = (n+2)! - 1.

    \sum_{r=1}^{n+1} r(r!) = \sum_{r=1}^{n} r(r!) + (n+1)*(n+1)! = (n+1)! - 1 +  (n+1)*(n+1)!.

    For the terms involving n take out a common factor of (n+1)! to get...

    (1+(n+1))*(n+1)! - 1 = (n+2)*(n+1)! - 1 = (n+2)! - 1. which is what we wanted. Hence by our inductive hypothesis the equality holds.
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  3. #3
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    Wow. Great answer, very clear. You probably won't beleive me but i was kind of halway there, i showed it held for n=1 but got confused half way through showing it holds for n+1!
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