Square of any odd integer is of the form where k is any integer.

Correct option is D

Given:

Let an odd integer be represented as a=2n+1, where n ∈ I (non-negative integer).

Formula:
a² = (2n + 1)² = 4n² + 4n + 1

Solution:
Since n ∈ I, consider two sub-cases:

Case I: When n is even →
Let n = 2p
Then,
a² = (2(2p) + 1)² = (4p + 1)²
= 16p² + 8p + 1
= 8(2p² + p) + 1 → of the form 8k + 1

Case II: When n is odd →
Let n = 2r+1
Then,
a² = (2(2r + 1) + 1)² = (4r + 3)²
= 16r² + 24r + 9
= 8(2r² + 3r + 1) + 1 → again of the form 8k + 1

Hence, from both cases:
Square of any odd integer = 8k + 1, where k is any integer.

Therefore, the correct option is (d) 8k + 1.