<s>[INST] Convert 2 + 5 +8+ . . . + (6n-1) = n(6n+1) to Sigma Notation 2 + 5 + 8 + . . . + (6n-1) = n(6n+1)
This is what I have so far. 
The sum of (3j-1) from j=1 to something I`m not sure of.
 [/INST] $$\sum\limits_{i=1}^{2n} (3i-1) = n(6n+1)$$
This will give you as terms of the series: \begin{align*}3(1)-1&=2\\3(2)-1&=5\\\ldots\\3(2n)-1&=6n-1\end{align*}
EDIT: the problem statement purported that $\sum_{i=1}^{2n} (3i-1) = n(6n-1)$. I will just prove below that this is wrong, and $\sum_{i=1}^{2n} (3i-1) = n(6n+1)$.
\begin{align*}
\sum\limits_{i=1}^{2n} (3i-1) &= \sum\limits_{i=1}^{2n} 3i + \sum\limits_{i=1}^{2n}-1
\\ &= \sum\limits_{i=1}^{2n} 3i + (-2n)
\\ &= 3\sum\limits_{i=1}^{2n} i + (-2n)
\end{align*}
Then we just use the summation formula: $\sum_{i=1}^{N}i = \frac{N(N+1)}{2}$.
\begin{align*}
\\ 3\sum\limits_{i=1}^{2n} i + (-2n) &= 3\left( \frac{2n(2n+1)}{2}\right) + (-2n)
\\ &= 3n(2n+1)-2n
\\ &= 6n^2+3n-2n
\\ &= 6n^2+n
\\ &= n(6n+1)
\end{align*}
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