The correct options are
A 3
C 9
4n−3n−1=(1+3)n−3n−1=(1+nC13+nC232+.........nCn3n)−3n−1
=(1+3n+nC232+........)−3n−1=nC232+nC333+............+nCn3n
=32(nC2+nC33+...........+nCn3n−2)=9× Integer.
Hence given number is divisible by 9. Hence it is also divisible by 3.