Again, start by writing down the recurrence relation when \(n = 1\text{.}\) This time, don't subtract the \(a_{n-1}\) terms to the other side:

\begin{equation*} a_1 = a_0 + 1\text{.} \end{equation*}

Now \(a_2 = a_1 + 2\text{,}\) but we know what \(a_1\) is. By substitution, we get

\begin{equation*} a_2 = (a_0 + 1) + 2\text{.} \end{equation*}

Now go to \(a_3 = a_2 + 3\text{,}\) using our known value of \(a_2\text{:}\)

\begin{equation*} a_3 = ((a_0 + 1) + 2) + 3\text{.} \end{equation*}

We notice a pattern. Each time, we take the previous term and add the current index. So

\begin{equation*} a_n = ((((a_0 + 1) +2)+3)+\cdots + n-1) + n\text{.} \end{equation*}

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*}