1 + 5 + 9 + …+(4n – 3) = n (2n – 1) for all natural number n.

Given; P(n) is 1 + 5 + 9 + …+(4n – 3) = n (2n – 1).

P(1) = 1 = 1(2−1) ; it’s true.

P(2) = 1 + 5 = 6 =2(4−1) ; it’s true.

P(3) = 1 + 5 + 9 = 15 = 3(6−1) ; it’s true.

Let P(k) be 1 + 5 + 9 + …+(4k – 3) = k (2k – 1) is true;

⇒ P(k+1) is 1 + 5 + 9 + …+(4k – 3) + (4(k+1)−3)

= k (2k – 1) + 4k + 1

= 2k² – k + 4k + 1

= (k+1)(2(k+1)−1)

⇒ P(k+1) is true when P(k) is true.

∴ By Mathematical Induction 1 + 5 + 9 + …+(4n – 3) = n (2n – 1) is true for all natural numbers n.