# Show that all Polygonal sequences generated by the Series of Arithmetic sequence with common difference d, d in ZZ are polygonal sequences that can be generated by a_n = an^2+bn+c?

Apr 9, 2018

${a}_{n} = {P}_{n}^{d + 2} = a {n}^{2} + {b}^{n} + c$
with a=d/2; b=(2-d)/2; c=0
${P}_{n}^{d + 2}$ is a polygonal series of rank, $r = d + 2$
example given an Arithmetic sequence skip counting by $d = 3$
you will have a $\textcolor{red}{p e n t a g o n a l}$ sequence:
${P}_{n}^{\textcolor{red}{5}} = \frac{3}{2} {n}^{2} - \frac{1}{2} n$ giving ${P}_{n}^{5} = \left\{1 , \textcolor{red}{5} , 12 , 22 , 35 , 51 , \cdots\right\}$

#### Explanation:

A polygonal sequence is constructed by taking the $n t h$ sum of an arithmetic sequence. In calculus, this would be an integration.
So the key hypothesis here is:
Since the arithmetic sequence is linear (think linear equation) then integrating the linear sequence will result in a polynomial sequence of degree 2.

Now to show this the case
${a}_{n} = \left\{1 , 2 , 3 , 4 , \cdots , n\right\}$
find the nth sum of ${S}_{n} = {\sum}_{i}^{i = n} {a}_{n}$
S_1 = 1; S_2 = 3, S_3=6, cdots
S_n = (a_1 + a_n)/2 n;
${a}_{n}$ is Arithmetic Sequence with
a_n= a_1 +d(n-1); a_1 = 1; d=1

${S}_{n} = \frac{1 + {a}_{n}}{2} n = \frac{\left(1 + 1 + \left(n - 1\right)\right)}{2} n = n \frac{n + 1}{2}$
${S}_{n} = {P}_{n}^{3} = \left\{1 , 3 , 6 , 10 , \cdots , \left(\frac{1}{2} {n}^{2} + \frac{1}{2} n\right)\right\}$
So with d = 1 the sequence is of the form ${P}_{n}^{3} = a {n}^{2} + b n + c$
with a = 1/2; b=1/2; c = 0

Now generalize for an arbitrary skip counter $\textcolor{red}{d}$, $\textcolor{red}{d} \in \textcolor{b l u e}{\mathbb{Z}}$ and ${a}_{1} = 1$:

${P}_{n}^{d + 2} = {S}_{n} = \frac{{a}_{1} + {a}_{1} + \textcolor{red}{d} \left(n - 1\right)}{2} n$
${P}_{n}^{d + 2} = \frac{2 + \textcolor{red}{d} \left(n - 1\right)}{2} n$
${P}_{n}^{d + 2} = \frac{\textcolor{red}{d}}{2} {n}^{2} + \left(2 - \textcolor{red}{d}\right) \frac{n}{2}$
Which is a general form ${P}_{n}^{d + 2} = a {n}^{2} + b n + c$
with a=color(red)d/2; b=(2-color(red)d)/2; c=0