Let the given statement be P(n), i.e., P (n) : 1 . 2 + 2 . 3 + 3 . 4 + ... + n . (n + 1) = $\left[ \frac{n(n+1)(n+2)}{2} \right]$ First we prove that the statement is true for n = 1. P (1) : 1 . 2 = $\frac{1(1+1)(1+2)}{3}$ ⇒ 2 = $\frac{2.3}{3}$ = 2 , which is true Assume P(k) is true for some positive integer k, i.e., 1 . 2 + 2 . 3 + 3 . 4 + ... + k . (k + 1) = $\left[ \frac{k(k+1)(k+2)}{2} \right]$ ... (i) We shall now prove that P(k + 1) is also true. For this we have to prove that 1 . 2 + 2 . 3 + 3 . 4 + ... + k . (k + 1) (k + 2) = $\frac{(k+1)(k+2)(k+3)}{3}$ L.H.S. = 1 · 2 + 2 · 3 + 3 · 4 + ... + k(k + 1) + (k + 1)(k + 2) = $\frac{k(k+1)(k+2)}{3}$ + (k + 1) (k + 2) [From (i)] = (k + 1) (k + 2) $\left[ \frac{k}{3} + 1 \right]$ = $\frac{(K+1)(k+2)(k+3)}{3}$ = 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.