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: Write out the series for $a_{n}$ to start with. We have that
$a_{n} = a_{n-1} + (n+1)\\
\quad = a_{n-2} + n + (n+1) \\
\quad = \ldots \\
\quad = a_{1} + 3 + 4 + \ldots + (n+1) \\
\quad = 2 + 3 + 4 + \ldots + (n+1) \\
\quad = \displaystyle \left(\sum_{i=1}^{n+1} i \right) - 1 \\
\quad = (n+1)(n+2)/2 - 1, \quad (\text{using the value for the sum of the integers between 1 and}\ n + 1) \\
\quad = (n^{2} + 3n)/2 + 1 -1 \\
\quad = (n^{2} + 3n)/2$ 
A: $a_n$ 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 $2a_n = n(n+3)$.
A: Hint: $a_n=a_{n-1}+(n+1)=a_{n-2}+n+(n+1)=\ldots=a_1+3+\ldots+(n+1)$.
A: Let $2\leq n$, then you can write 
$$a_{2}-a_{1}=2+1$$
$$a_{3}-a_{2}=3+1$$
$$...$$
$$a_{n}-a_{n-1}=n+1.$$
Now addition of all this equality implies that 
$$a_{n}-a_{1}=(2+3+..+n)+(n-1),$$
$$a_{n}=(2+3+..+n)+(n+1)-1,$$
we know that $1+2+..+n=\frac{n^{2}+n}{2}$, so
$$a_{n}=\frac{(n+1)(n+2)}{2}-1$$
