We have \(2 = 2\text{,}\) \(7 = 2+5\text{,}\) \(15 = 2 + 5 + 8\text{,}\) \(26 = 2+5+8+11\text{,}\) and so on. The terms in the sums are given by the arithmetic sequence \(b_n = 2+3n\text{.}\) In other words, \(a_n = \sum_{k=0}^n (2+3k)\text{.}\) To find the closed formula, we reverse and add. We get \(a_n = \frac{(4+3n)(n+1)}{2}\) (we have \(n+1\) there because there are \(n+1\) terms in the sum for \(a_n\)).