# Find formulas for $a_n$ for hexagonal arrangements

#### Robb

Problem 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!

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#### fresh_42

Mentor
2018 Award
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

Mentor
2018 Award
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*}
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!

"Find formulas for $a_n$ for hexagonal arrangements"

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