Regrouping terms, we notice that \(a_n\) is just \(a_0\) plus the sum of the integers from \(1\) to \(n\text{.}\) So, since \(a_0 = 4\text{,}\)

\begin{equation*} a_n = 4 + \frac{n(n+1)}{2}\text{.} \end{equation*}