\((1, 2, 4, 7, 11, 16, 22, \ldots)\text{.}\) Note that if subtract 1 from each term, we get the sequence \((T_n)\text{.}\) So we have \(b_n = T_n + 1\text{.}\) Therefore a closed formula is \(b_n = \frac{n(n+1)}{2} + 1\text{.}\) A quick check of the first few \(n\) confirms we have it right.
in-context