The correct options are
B 0, if n is odd
C (−1)n/2 nCn/2, if n is even
Given , C20−C21+C22−......(−1)n×C2n
Case I : When n is odd, i.e. n=2m+1,
S=C20−C21+C22−.....+(−1)2m×C22m+(−1)2m+1×C2m+1
S=C20−C21+C22−.....+C22m−C22m+1
=(C20−C22m+1)−(C21−C22m)+....+(−1)m(C2m−C2m+1)
But
2m+1C0=2m+1C2m+1;2m+1C1=2m+1C2m etc.
Therefore, S=0
Case II: When n is even, i.e. n=2m.
S=C20−C21+C22−......+(−1)2m×C22m
S=C20−C21+C22......+C22m
Consider the coefficient of the constant term in the following expression.
[C0−C1x+C2x2−.....+(−1)2m×C2mx2m][C0+C11x+C21x2+....+C2m1x2m]
=(1−x)2m(1+1x)2m
Coefficient of x2m in (1−x)2m(1+x)2m
= Coefficient of x2m in (1−x2)2m
=(−1)m2mCm=(−1)n/2nCn/2