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)

Your cleary not see the answer is 25