Hello, AfterShock!

Quote:

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 firstn"triangular" numbers.

Since it starts with n = 3, thenth 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