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.