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Proof By induction Sigma notation...Please help 
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#1
Dec1811, 07:42 PM

P: 4

1. The problem statement, all variables and given/known data
2n1 Sigma (3i+1) = n(6n1) 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(6m1) + (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(i1)+1) i=1 and still couldn't prove it. Thanks for your time 


#2
Dec1811, 09:15 PM

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Now, assume that it holds for n = m, (where m ≥ 1), so you assume the following is true. [itex]\displaystyle\sum_{i=0}^{2m1}(3i+1)= m(6m1)\,.[/itex]With that assumption, you now need to show that this formulation is true for n = m+1. In this case, 2n1 = 2(m+1)1 = 2m+1, and n(6n1)= (m+1)(6m+5)= 6m^{2}+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}^{ 2m1}(3i+1)\,.[/itex] 


#3
Dec1811, 11:27 PM

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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