First, if you look at the differences between terms, you get a sequence of differences: \(1,4,7,10,13, \ldots\text{,}\) which is an arithmetic sequence. Written another way:

\begin{align*} a_0 \amp = 2\\ a_1 \amp = 2+1\\ a_2 \amp = 2+1+4\\ a_3 \amp = 2+1+4+7 \end{align*}

and so on. We can write the general term of \((a_n)\) in terms of the arithmetic sequence as follows:

\begin{equation*} a_n = 2 + 1 + 4 + 7 + 10 + \cdots + (1+3(n-1)) \end{equation*}

(we use \(1+3(n-1)\) instead of \(1+3n\) to get the indices to line up correctly; for \(a_3\) we add up to 7, which is \(1+3(3-1)\)).

We can reverse and add, but the initial 2 does not fit our pattern. This just means we need to keep the 2 out of the reverse part:

\(a_n =\)	\(2\)	\(+\)	\(1\)	\(+\)	\(4\)	\(+ \cdots +\)	\(1+3(n-1)\)
\(+ \quad a_n =\)	\(2\)	\(+\)	\(1+3(n-1)\)	\(+\)	\(1+3(n-2)\)	\(+ \cdots +\)	\(1\)
\(2a_n =\)	\(4\)	\(+\)	\(2+3(n-1)\)	\(+\)	\(2+3(n-1)\)	\(+ \cdots +\)	\(2+3(n-1)\)

Not counting the first term (the 4) there are \(n\) summands of \(2+3(n-1) = 3n-1\) so the right-hand side becomes \(4+(3n-1)n\text{.}\)

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.