Let the given statement be P(n), i.e., P (n) : $1^2+3^2+5^2$ + ... + $(2n-1)^2$ = $\frac{n(2n-1)(2n+1)}{3}$ First we prove that the statement is true for n = 1. P (1) : $1^2$ = $\frac{1\cdot(2\cdot1-1)(2\cdot1+1)}{3}$ ⇒ 1 = $\frac{1\cdot1\cdot3}{3}$ = 1 which is true Assume P(k) is true for some positive integer k, i.e., $1^2+3^2+5^2$ + ... + $(2k-1)^2$ = $\frac{k(2k-1)(2k+1)}{3}$ ... (i) Now we shall prove that P(k + 1) is also true. For this we have to prove that $1^2+3^2+5^2$ + ... + $(2k-1)^2 + (2(k+1)-1)^2$ = $\frac{(k+1)(2(k+1)-1)(2(k+1)+1)}{3}$ L.H.S. = $1^2+3^2+5^2$ + ... + $(2k-1)^2 + (2(k+1)-1)^2$ = $\frac{k(2k-1)(2k+1)}{3}$ + $[2(k+1)-1]^2$ [From (i)] = $\frac{k(2k-1)(2k+1)}{3} + [2k+1]^2$ = (2k + 1) $\left[\frac{k(2k-1)}{3} + (2k+1)\right]$ = (2k + 1) $\left[\frac{2k^2 - k + 6k + 3}{3}\right]$ = (2k + 1) $\left(\frac{2k^2 + 5k + 3}{3}\right)$ = $\frac{(2k+1)(2k+3)(k+1)}{3}$ = $\frac{(k+1)(2(k+1)-1)(2(k+1)+1)}{3}$ = R.H.S. Thus P(k + 1) is true, whenever P(k) is true. Hence, by the principle of mathematical induction, P(n) is true ∀ n ∈ N.