Question

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`?

Answer 1

`a_n=P_n^(d+2)= an^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 `color(red)(pentagonal)` sequence:
`P_n^color(red)5= 3/2n^2-1/2n` giving `P_n^5={1, color(red)5, 12, 22,35,51, cdots }`

Explanation:

A polygonal sequence is constructed by taking the `nth` 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
Start with a natural sequence (skip counting by starting with 1)
`a_n = {1, 2,3,4, cdots, n}`
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 = (1+a_n)/2 n= [(1+1+(n-1))]/2n = n(n+1)/2`
`S_n = P_n^3 = {1, 3, 6, 10, cdots, (1/2n^2+1/2n)}`
So with d = 1 the sequence is of the form `P_n^3 = an^2+bn+c`
with `a = 1/2; b=1/2; c = 0`

Now generalize for an arbitrary skip counter `color(red)d`, `color(red)d in color(blue)ZZ` and `a_1 = 1`:

`P_n^(d+2) = S_n = (a_1+a_1+color(red)d(n-1))/2 n`
`P_n^(d+2)= (2+color(red)d(n-1))/2 n`
`P_n^(d+2) =color(red)d/2n^2+(2-color(red)d)n/2`
Which is a general form `P_n^(d+2) =an^2+bn+c`
with `a=color(red)d/2; b=(2-color(red)d)/2; c=0`