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Find formulas for ##a_n## for hexagonal arrangements

  • Thread starter Robb
  • Start date
207
6
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!
 

fresh_42

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2018 Award
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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?
 
207
6
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
Insights Author
2018 Award
10,039
6,780
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*}
 
207
6
Right, I forgot about that! Thanks!
 

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