by jesusjones
 P: 4 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
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 Quote by jesusjones 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 ...
Hello jesusjones. Welcome to PF.

Now, assume that it holds for n = m, (where m ≥ 1), so you assume the following is true.
$\displaystyle\sum_{i=0}^{2m-1}(3i+1)= m(6m-1)\,.$
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.
$\displaystyle\sum_{i=0}^{2m+1}(3i+1) = (m+1)(6m+5)\,.$
Here's a hint:
$\displaystyle\sum_{i=0}^{2m+1}(3i+1)=(3(2m)+1)+(3(2m+1)+1)+\sum_{i=0}^{ 2m-1}(3i+1)\,.$
 P: 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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