The correct option is C 2n.(n+1)
Consider the following
(1+x2)n=1+nC1x2+nC2x4+nC3x6...nCnx2n
Multiplying both sides with x, we get
x(1+x2)n=x+nC1x3+nC2x5+nC3x7...nCnx2n+1
Differentiating both sides with respect to x, we get
[(1+x2)n+nx(2x)(1+x2)n−1]x=1=1+3nC1+5nC2+...(2n+1)nCn
Hence
1+3nC1+5nC2+...(2n+1)nCn
=[(1+x2)n+nx(2x)(1+x2)n−1]x=1
=2n+2n(2n−1)
=2n+n2n
=2n(n+1)