Question 17463
the first 20 multiples of 3 are:
3, 6, 9, 12, 15, etc
we need the sum
3 + 6 + 9 + 12 + 15 + ....

Note 3 = 3*1
     6 = 3*2
     9 = 3*3
     12 = 3*4
     15 = 3*5
       etc.

So, we need:
3*1 + 3*2 + 3*3 + 3*4 + 3*5 + ... +3*20
= 3[1 + 2 + 3 + 4 + 5 + ... + 20]
= 3[210]
= 630