Proof By induction Sigma notation Please help

  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. SammyS

    SammyS 9,036
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    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. 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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