Explicit Formula for a Recurrence Relation for {2, 5, 9, 14, ...} (Chartrand Ex 6.46[b]) 
Consider the sequence $a_1 = 2, a_2 = 5, a_3 = 9, a_4 = 14,$ etc...
  (a) The recurrence relation is: $a_1 = 2$ and $a_n = a_{n - 1} + (n + 1) \; \forall \;n \in [\mathbb{Z \geq 2}]$.
  (b) Conjecture an explicit formula for $a_n$. (Proof for conjecture pretermitted here)

I wrote out some $a_n$ to compass to cotton on to an idea or pattern. It seems bootless.
$\begin{array}{cc}
 a_2 = 5 & a_3 = 9 & a_4 = 14 & a_5 = 20 & a_6 = 27\\
\hline \\
5 = 2 + (2 + 1) & 9 = 5 + (3 + 1) & 14 = 9 + (4 + 1) & 20 = 14 + (5 + 1) & 27 = 20 + (7 + 1) \\
\end{array}$
The snippy answer only says $a_n = (n^2 + 3n)/2$. Thus, could someone please explicate the (missing) motivation or steps towards this conjecture? How and why would one envision this? 

$a_n$ is equal to $2 + 3 + \ldots + (n+1)$. You can apply reasoning to $a_n$ by taking two copies of $a_n$, one with sum of terms in forward direction, and one with sum of terms in reverse direction, and you will get $n$ pairs of terms each of which add up to $n+3$. So $2a_n = n(n+3)$.
