The correct option is D 150a
Substituting x=1 is (1+x+x2+x3)100, we get
a=4100
Differentiating (1+x+x2+x3)100, with respect to x, we get
100(1+x+x2+x3)99(1+2x+3x2)
=∑rar
Substituting x=1, in the above equation, we get
∑rar=100(499)(6)
=600(499)
=600(4100)4
=150(499)
=150(a)