The sume of the first 100 positive multiples of 4 is _ more than the sum of the first 100 positive multiples of 3.

Compare the sums for a second: 

4+8+12+16+20+24

3+6+9+12+15+18

Sum of the first 100 positive integers, so 5050 or (D).

The arithmetic sum formula shows that:

\(\text{sum} = 100(\frac{4+100}{2}) = 5200\)

\(\text{sum} = 100(\frac{3+100}{2}) = 5150\)

The difference is 5200 - 5150 = 50.

tertre is correct

4 + 8 + 12 + 16 +.....+  400

3 + 6  + 9  + 12 +.....+  300

The difference in the series is

1 + 2 + 3 + 4 + ....+ 100   =  the sum of the first 100 positive integers =   100 (101) / 2  =  50 * 101 = 5050