I will prove here a theorem I have not seen before:
The sum of an arithmetic progression with common difference k
is the sum of k triangular numbers of two sizes.
Consider the arithmetic progression h, h + k, h + 2k, h + 3k, ..., h+(n-1)k which has n terms.
The summation of this is the series: hn + k(1 + 2 + 3 + ...+ (n-1))
= hn + kn(n-1)/2 = [kn^2 + (2h - k)n)]/2
This formula can be expressed as (k-h)n(n-1)/2 + hn(n+1)/2
= (k-h)T(n-1) + hT(n).
That is as the sum of k triangles, h of side n and (k-h) of side (n-1).
In the case when h= 0 or h = k we get k triangles all of the same size.
When h = 1 and k = 3 we get the sequence 1, 4, 7, 10, 13, 16, 19, 22, 25, ...
its summation is the series: 1, 5, 12, 22, 35, 51, 70, 92, 117, ...
which is traditionally known as the 'pentagonal numbers'.
When h = 2 and k = 3 we get the sequence 2, 5, 8, 11, 14, 17, 20, 23, 26, ...
its summation is the series: 2, 7, 15, 26, 40, 57, 77, 100, 126, ...
which are known as the 'pentagonal numbers of the second kind'.
When h = 0 (or 3) we get the sequence 0, 3, 6, 9, 12, 15, 18, 21, 24, 27, ...
its summation is the series 0, 3, 9, 18, 30, 45, 63, 84, 108, 135, ...
which are triple triangle numbers.
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