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Proof By induction Sigma notation Please help

  1. Dec 18, 2011 #1
    1. The problem statement, all variables and given/known data

    2n-1
    Sigma (3i+1) = n(6n-1)
    i=0

    prove for all positive n


    2. Relevant equations



    3. The attempt at a solution

    It holds true for n=1

    5=5

    then P: m+1

    2m+1
    Sigma(3i+1) = (m+1)(6(m+1)-1) or 6m^2 + 11m + 5
    i=0

    then 2m+1
    Sigma(3i+1) = m(6m-1) + (3(m+1)+1) + (3(m+2)+1) = 6m^2 + 5m + 11
    i=0

    I just cannot figure this out and it is driving me crazy. Please help clarify things. I even tried changing the first part to
    2m
    Ʃ(3(i-1)+1)
    i=1

    and still couldn't prove it.

    Thanks for your time
     
  2. jcsd
  3. Dec 18, 2011 #2

    SammyS

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    Hello jesusjones. Welcome to PF.

    Now, assume that it holds for n = m, (where m ≥ 1), so you assume the following is true.
    [itex]\displaystyle\sum_{i=0}^{2m-1}(3i+1)= m(6m-1)\,.[/itex]​
    With that assumption, you now need to show that this formulation is true for n = m+1. In this case, 2n-1 = 2(m+1)-1 = 2m+1, and n(6n-1)= (m+1)(6m+5)= 6m2+11m+5 . In other words, show that the following can be derived from the above.
    [itex]\displaystyle\sum_{i=0}^{2m+1}(3i+1) = (m+1)(6m+5)\,.[/itex]​

    Here's a hint:
    [itex]\displaystyle\sum_{i=0}^{2m+1}(3i+1)=(3(2m)+1)+(3(2m+1)+1)+\sum_{i=0}^{2m-1}(3i+1)\,.[/itex]​
     
  4. Dec 18, 2011 #3
    Thank you SammyS for your reply your hint lead me to the proper proof. I was having a hard time figuring out why it would be (3(2m)+1) and (3(2m+1)+1) instead of just (3(m+1)+1) ect.... But i think i get it now.

    Thanks very much
     
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