Let $P(n): 1.2 .3+2.3 .4+\ldots \ldots \ldots+n(n+1)(n+2)$
$=\frac{n(n+1)(n+2)(n+3)}{4} \quad \ldots(i)$
For $n=1, \mathrm{LHS}=1.2 .3=6$ and
RHS $=\frac{1(1+1)(1+2)(1+3)}{4}=\frac{1 \times 2 \times 3 \times 4}{4}=6$
$\therefore$ LHS $=$ RHS i.e., $P(1)$ is true.
Let $P(k)$ is true.
$\therefore P(k): 1.2 .3+2.3 .4+\ldots \ldots \ldots+k(k+1)(k+2)$
$=\frac{k(k+1)(k+2)(k+3)}{4} \quad \ldots(ii)$
We have to prove that
$P(k+1)$ is true whenever $\mathrm{P}(\mathrm{k})$ is true.
$\therefore P(k+1):\left[\begin{array}{ll}1.2 .3 & +2.3 .4+3.4 .5+\ldots \ldots \ldots \ldots \\ & +k(k+1)(k+2)]+(k+1)(k+2)(k+3)\end{array}\right.$
$=\frac{k(k+1)(k+2)(k+3)}{4}+(k+1)(k+2)(k+3)$
$=(k+1)(k+2)(k+3)\left(\frac{k}{4}+1\right) \quad[$ from (ii)]
$\begin{aligned}
&=\frac{(k+1)(k+2)(k+3)(k+4)}{4} \\
&=\frac{(k+1)(k+2)(k+3)(k+4)}{4} \quad \ldots(iii)
\end{aligned}$
$\therefore \quad P(n)$ is true for $n=k+1$ i.e., $P(k+1)$ is true whenever $P(k)$ is true.
$\therefore$ By Principle of Mathematical Induction,
$\therefore \quad P(n)$ is true for all natural numbers $n$.