Hello, AfterShock!
n = 3 . → . .1
n = 4 . → . .4
n = 5 . → . 10
n = 6 . → . 20
n = 7 . → . 35
n = 8 . → . 56
n = 9 . → . 84 . . . and so forth.
I need the sum for n ≥ 3.
I recognized the pattern . . .
S .= .1 + (1 + 2) + (1 + 2 + 3) + (1 + 2 + 3 + 4) + ...
This is the sum of the first n "triangular" numbers.
Since it starts with n = 3, the nth term is: ½(n - 2)(n - 1)
We have: .S .= .∑ ½(n - 2)(n - 1) .= .½ ∑(n² - 3n + 2) .= .½(∑n² - 3∑n + 2∑1)
. . . . . . . .S .= .½[n(n + 1)(2n + 1)/6 - 3n(n + 1)/2 + 2n]
which simplifies to: . S .= .n(n - 1)(n - 2)/6