Consider the equation. (1+x+x2)n=c0+c1x+c2x2+⋅s Substitute ‌‌
1
x
for x in the above equation. ‌‌
(1−x+x2)n
x2n
=‌‌
c0x2n−c1x2n−1+c2x2n−2+c3x2n−3+.....
x2n
(1−x+x2)n=c0x2n−c1x2n−1+c2x2n−3+.... Now c0x2n+c1x2n−1+c2x2n−3+....=‌ Coeff of ‌x2n+1‌ in ‌(1+x+x2)n(1−x+x2)n =‌ Coeff of ‌x2n+1in((1+x2)2−x2)n =‌ Coeff of ‌x2n+1in(1+x4+x2)n =0 since the coefficient of xk in the expansion if (1+x4+x2)n exist if k is even other it will be zero and 2n+1 is odd