What is the sum of the multiples of 3 between 3 and 999, inclusive?

3 + 6 + 9 + 12 + ... + 999

= 3 (1 + 2 + 3 + 4 + ... + 333)

= 3 (Summation of r from 1 to 333)

= 3 ((1 + 333) (333/2))

= 3 (334) (333/2)

= 3 (167) (333)

= 166833

Edit: Opps, I forgot that you ask about an equation.

The sum of 1 + 2 + 3 + 4 + ... + n, where n is some integer, is:

(First term + Last term) multiplied by (Total number of terms / 2)

= (1 + n) (n / 2)