The correct options are
A 0 if n is odd
B (−1)n/2(n2+1)n!(n2)!(n2)!
When n is odd, we take n=2m−1, so that
S=C20−C21+C22−......(−1)2m−1(2m)C22m−1 ........(1)
Using 2m−1Cr=2m−1C2m−1−r, we rewrite (1) as
S=−2mC20+(2m−1)C21+....+C22m−1 .......(2)
Adding (1) and (2), we get
2S=(1−2m)[C20−C21+C22−......(−1)2m−1C22m−1]
(1−2m)(0)=0⇒S=0
When n is even, we take n=2m
S=C20−C21+C22−......(−1)2m−1(2m)C2m−1+(−1)2m(2m+1)C22m .......(3)
Using 2mCr=2mC2m−r, we rewrite (3) as
S=(2m+1)C20−(2m)C21+(2m−1)C22−.....+(−1)2m−1C22m ........(4)
Adding (3) and (4), we get
2S=(2m+2)[C20−C21+C22−......(−1)2mC22m]
⇒S=(m+1)(−1)m(2mCm)
=(−1)n/2(n2+1)n!(n2)!(n2)!