Finally, solving for \(a_n\) we get

\begin{equation*} a_n = \d \frac{4+(3n-1)n}{2}\text{.} \end{equation*}

Just to be sure, we check \(a_0 = \frac{4}{2} = 2\text{,}\) \(a_1 = \frac{4+2}{2} = 3\text{,}\) etc. We have the correct closed formula.