The correct option is A zero
(1−x)n=C0−C1x+C2x2+....+(−1)nCnxn ....(1)
(1−x5)n=C0−C1x5+C2x10+....+(−1)nCnx5n
Differentiating wrt x
−5nx4(1−x5)n−1=−5C1x4+10C2x9+15C2x14+....+(−1)nCnx5n
Multiplying (1) by 3 and then adding with (2):
3(1−x)n−5nx4(1−x5)n−1=3.nC0−8nC1x4+13nC2x9−18nC3x14....
now, put x=0
0=3nC0−8nC1+13nC2−18nC3....