Let the given statement be P(n), i.e.,
P(n) : 1⋅ 2 ⋅ 3 + 2 ⋅ 3⋅ 4+....+ n(n + 1)(n + 2) =

n(n+1)(n+2)(n+3)

4

First we prove that the statement is true for n = 1.
P (1) : 1 . 2 . 3 =

1(1+1)(1+2)(1+3)

4

⇒ 6 =

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)
=

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)
=

(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)
=

k(k+1)(k+2)(k+3)

4

+ (k + 1) (k + 2) (k + 3) [From (i)]
= (k + 1) (k + 2) (k + 3) [

k

4

+1] = (k + 1) (k + 2) (k + 3) [

k+4

4

] = 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.