Find formulas for ##a_n## for hexagonal arrangements

[Robb]

Homework Statement

Let $S_n$be the hexagonal arrangements consisting of n rings of dots for $n \in {1, 2, 3}$. Let $a_n$ be the number of dots in $S_n$. Find formulas for $a_n$ and $\sum_{k=1}^n a_k$.

Relevant Equations

$a_1=1, a_2=7, a_3=19$

$a_n = a_{n-1} + 6(n-1)$
$=a_{n-2} + 6(n-1) + 6(n-2)$
$=a_{n-3} + 6(n-1) + 6(n-2) + 6(n-3)$
$\vdots$
$=1 + 6 [(n-1) + (n-2) + (n-3) + \cdots + 1]$
$= 1 + 3n(n-1)$

I'm not sure how to get to the last line from the second to the last line. Please advise. Thanks!

 

[fresh_42]

The formula for the sum of $n$ consecutive numbers is $\sum_{k=1}^n k = \frac{1}{2}n(n+1)\,.$
Do you know how this can be proven?

 

[Robb]

Inductively?
$\sum_{k=1}^n k = \frac 1 2 n(n+1)$
$\sum_{k=1}^{n+1} k = \frac 1 2 (n+1)(n+2)$
$\sum_{k=1}^n k = \frac 1 2 n(n+1) + (n+1)$
$= \frac 1 2 [n(n+1) +2(n+1)]$
$= \frac 1 2 [n^2 + 3n + 2]$
$= \frac 1 2 [(n+1)(n+2)]$
as required

 

[fresh_42]

Inductively?
$\sum_{k=1}^n k = \frac 1 2 n(n+1)$
$\sum_{k=1}^{n+1} k = \frac 1 2 (n+1)(n+2)$
$\sum_{k=1}^n k = \frac 1 2 n(n+1) + (n+1)$
$= \frac 1 2 [n(n+1) +2(n+1)]$
$= \frac 1 2 [n^2 + 3n + 2]$
$= \frac 1 2 [(n+1)(n+2)]$
as required

Or you can use Gauss' trick:
\begin{align*}
1 &\quad \quad 2 & 3 & \ldots & \frac{n}{2} \\
n &\quad n-1 & n-2 & \ldots & \frac{n}{2}+1 \\
+ &&&&\\
\hline \\
n+1&\quad n+1 &n+1 & \ldots & n+1 \\
=\frac{n}{2} \cdot (n+1)&&&&
\end{align*}

 

[Robb]

Right, I forgot about that! Thanks!