Let the given statement be P(n), i.e., P(n) : 1⋅ 2 ⋅ 3 + 2 ⋅ 3⋅ 4+....+ n(n + 1)(n + 2) = $\frac{n(n+1)(n+2)(n+3)}{4}$ First we prove that the statement is true for n = 1. P (1) : 1 . 2 . 3 = $\frac{1(1+1)(1+2)(1+3)}{4}$ ⇒ 6 = $\frac{2.3.4}{4}$ = 6, which is true Assume P(k) is true for some positive integer k, i.e., 1·2·3 + 2·3·4 + ..... + k(k + 1)(k + 2) = $\frac{k(k+1)(k+2)(k+3)}{4}$ ... (i) We shall now prove that P(k + 1) is also true. For this we have to prove that 1·2·3 + 2·3·4 + .. + k(k + 1)(k + 2) + (k + 1)(k + 2) (k + 3) = $\frac{(k+1)(k+2)(k+3)(k+4)}{4}$ L.H.S. = 1·2·3 + 2·3·4 + ..... + k(k + 1)(k + 2) + (k + 1)(k + 2)(k + 3) = $\frac{k(k+1)(k+2)(k+3)}{4}$ + (k + 1) (k + 2) (k + 3) [From (i)] = (k + 1) (k + 2) (k + 3) $\left[ \frac{k}{4}+1 \right]$ = (k + 1) (k + 2) (k + 3) $\left[ \frac{k+4}{4} \right]$ = R.H.S. Thus P(k + 1) is true, whenever P(k) is true. Hence, by the principle of mathematical induction, the statement P(n) is true ∀ n ∈ N.