It is given that, (1+x+x2)=a0+a1x+a2x2+...+a2nx2n⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(I) Substitute 1 for x in equation (I) (1+1+12)n=a0+a11+a212+⋅s+a2n12n 3n=a0+a1+a2+⋅s+a2n Substitute i2 for x in equation (I) 1=a0−a1+a2+...+a2n Add both the equation. 3n+1=2(a0+a2+⋅s+a2n) (a0+a1+a2+⋅s+a2n)=
1
2
(3n−1) Now, (a1+a3+a5+⋅s+a2n−1)=
1
2
(3n−1) Differentiate equation (I) with respect to x n(1+x+x2)n−1(1+2x)=a1+2a2x+⋅s+2na2nx2n−1 At x=1, the equation becomes n⋅3n=a1+2a2+⋅s2na2n Therefore, the correct option is c