The correct option is D 2q+1
Let ′a′ be a given positive integer. On dividing ′a′ by 2, let q be the quotient and r be the remainder. Then, by Euclid's division algorithm, we have
a=2q+r, where 0≤r<2⇒a=2q+r,wherer=0 or r=1⇒a=2q or 2q+1when a=2q+1 for some integer q,then clearly ′a′ is odd.