Let the given statement be P(n), i.e., P (n) : 1 . 3 + 3 . 5 + 5 . 7 + ... + (2n - 1) (2n + 1) = $\frac{n(4n^2+6n-1)}{3}$ First we prove that, the statement is true for n = 1. P (1) : 1 . 3 = $\frac{1\cdot(4\cdot(1)^2+6\cdot1-1)}{3}$ ⇒ 3 = $\frac{4+6-1}{3}$ = $\frac{9}{3}$ = 3 , which is true Assume P(k) is true for some positive integer k, i.e., 1 . 3 + 3 . 5 + 5 . 7 + ... + (2k - 1) (2k + 1) = $\frac{k(4k^2+6k-1)}{3}$ ... (i) We shall now prove that P(k + 1) is also true. For this we have to prove that 1·3 + 3·5 + 5·7 + ........ + (2k – 1)(2k + 1) + (2k + 1)(2k + 3) = $\frac{(k+1)[4(k+1)^2+6(k+1)-1]}{3}$ L.H.S. = 1·3 + 3·5 + 5·7 + ..... + (2k – 1)(2k + 1) + (2k + 1)(2k + 3) = $\frac{k(4k^2+6k-1)}{3}$ + (2k + 1) (2k + 3) [From (i)] = $\frac{4k^3+6k^2-k+3(4k^2+8k+3)}{3}$ = $\frac{4k^3+6k^2-k+12k^2+24k+9}{3}$ = $\frac{4k^3+18k^2+23k+9}{3}$ ... (ii) Also, R.H.S. = $\frac{(k+1)[4(k+1)^2+6(k+1)-1]}{3}$ = = $\frac{(k+1)[4k^2+8k+4+6k+5]}{3}$ = $\frac{(k+1)[4k^2+14k+9]}{3}$ = $\frac{4k^3+14k^2+9k+4k^2+14k+9}{3}$ = $\frac{4k^3+18k^2+23k+9}{3}$ ... (iii) From (ii) and (iii), we get L.H.S. = 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.