1. Dec 18, 2011

### 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

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.

2. Dec 18, 2011

### SammyS

Staff Emeritus
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:
$\displaystyle\sum_{i=0}^{2m+1}(3i+1)=(3(2m)+1)+(3(2m+1)+1)+\sum_{i=0}^{2m-1}(3i+1)\,.$​

3. Dec 18, 2011

### jesusjones

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