Prove the following by using the principle of mathematical induction for all n ∈ N: 1.2.3 + 2.3.4 + … + n(n + 1) (n + 2)

Let the given statement be P(n), i.e.,

$\mathrm{P}(n): 1.2 .3+2.3 .4+\ldots+n(n+1)(n+2)=\frac{n(n+1)(n+2)(n+3)}{4}$

For n = 1, we have

$P(1): 1.2 .3=6=\frac{1(1+1)(1+2)(1+3)}{4}=\frac{1.2 .3 .4}{4}=6$, which is true.

Let P(k) be true for some positive integer k, i.e.,

$1.2 .3+2.3 .4+\ldots+k(k+1)(k+2)=\frac{k(k+1)(k+2)(k+3)}{4}$   $\ldots$ (i)

We shall now prove that P(k + 1) is true.

Consider

$1.2 .3+2.3 .4+\ldots+k(k+1)(k+2)+(k+1)(k+2)(k+3)$

$=\{1.2 .3+2.3 .4+\ldots+k(k+1)(k+2)\}+(k+1)(k+2)(k+3)$

$=\frac{k(k+1)(k+2)(k+3)}{4}+(k+1)(k+2)(k+3) \quad[$ Using (i) $]$

$=(k+1)(k+2)(k+3)\left(\frac{k}{4}+1\right)$

$=\frac{(k+1)(k+2)(k+3)(k+4)}{4}$

$=\frac{(k+1)(k+1+1)(k+1+2)(k+1+3)}{4}$

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.