The correct option is C 5
(1+3√2x)9+(1−3√2x)9
Applying binomial theorem for expansion of the following
=1+(3√2x)(9C1)+(3√2x)2(9C2)...(9C9(3√2x)9)
+1−(3√2x)(9C1)+(3√2x)2(9C2)...−(9C9(3√2x)9)
=2[1+(3√2x)2(9C2)+(3√2x)4(9C4)...(3√2x)8(9C8)]
The powers of 3√2x are in A.P
0,2,4,6,8
Hence, including the term 1orx0 there are 9+12 terms.
Hence, there are 5 terms,