(d) 2n + 1
Every positive odd integer is of the form 2n + 1
Proof:
Let n be the given positive odd integer.
On dividing n by 2, let q be the quotient and r be the remainder.
On applying Euclid's algorithm, we get:
n = 2q + r, where 0 ≤ r ≤ 2
⇒ n = 2q + r, where r = 0, 1
⇒ n = 2q and n = 2q + 1
But,
n = 2q = even
Thus, when n is odd, it is in the form of n = 2q + 1.