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#?

1 Answer

Answer:

#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#