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Induction Proof

  1. Aug 13, 2009 #1
    1. The problem statement, all variables and given/known data
    Prove that for n [tex]\geq[/tex] 1
    n
    [tex]\sum[/tex] m[tex]^{2}[/tex] = (1/6)*n(n+1)(2n+1)
    m=1


    2. Relevant equations



    3. The attempt at a solution

    Base Case:
    n=1

    (1)^2 = (1/6)*(6)
    1=1

    Inductive Step:
    Assume n=k
    k
    [tex]\sum[/tex] m^2 = (1/6)*k(k+1)(2k+1) is true
    m=1


    Let n=k+1

    k+1
    [tex]\sum[/tex] m^2 = (1/6)*(k+1)((k+1)+1)(2(k+1)+1)
    m=1

    Any hints to prove this? I've tried simplifying the last equation but I haven't had any luck
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 13, 2009 #2

    Mark44

    Staff: Mentor

    Split your last summation into one running from m = 1 to k plus the single term when m = k + 1. For the first summation, use what you have from your induction step.
     
  4. Aug 13, 2009 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Well, this is what you want to get.

    Recognise that
    [tex]\sum_{m=0}^{k+1} m^2= \sum_{m=0}^k m^2+ (k+1)^2[/tex]

    If
    [tex]\sum_{m=0}^k m^2= (1/6)*k(k+1)(2k+1)[/tex]
    then that becomes
    [tex]\sum_{m=0}^{k+1} m^2= (1/6)*k(k+1)(2k+1)+ (k+1)^2[/tex]
    That is what you must show is equal to
    [tex](1/6)*(k+1)((k+1)+1)(2(k+1)+1)[/tex]
     
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