Let $P(n): \frac{1}{1.2 .3}+\frac{1}{2.3 .4}+\frac{1}{3.4 .5}+\ldots \ldots \ldots$ $+\frac{1}{n(n+1)(n+2)}=\frac{n(n+3)}{4(n+1)(n+2)}$
Putting $n=1$, LHS $=\frac{1}{1.2 .3}=\frac{1}{6}$;
RHS $=\frac{1(1+3)}{4(1+1)(1+2)}=\frac{4}{4.2 .3}=\frac{1}{6}$
$\therefore$ LHS $=$ RHS $\quad \therefore P(n)$ is true for $n=1$
Assuming $P(n)$ is true for $n=k$
i.e., $P(k): \frac{1}{1.2 .3}+\frac{1}{2.3 .4}+\frac{1}{3.4 .5}+\ldots \ldots \ldots$
$+\frac{1}{k(k+1)(k+2)}=\frac{k(k+3)}{4(k+1)(k+2)} \quad \ldots(i)$
We have to prove that $P(k+1)$ is true. whenever $P(k)$ is true.
$\therefore P(k+1): \frac{1}{1.2 .3}+\frac{1}{2.3 .4}+\frac{1}{3.4 .5}+\ldots \ldots \ldots$
$+\frac{1}{k(k+1)(k+2)}+\frac{1}{(k+1)(k+2)(k+3)}$
$=\frac{k(k+3)}{4(k+1)(k+2)}+\frac{1}{(k+1)(k+2)(k+3)} \quad [from(i)]$
$=\frac{1}{(k+1)(k+2)}\left[\frac{k(k+3)}{4}+\frac{1}{k+3}\right]$
$=\frac{1}{(k+1)(k+2)} \times \frac{k(k+3)^2+4}{4(k+3)}$
$=\frac{1}{4(k+1)(k+2)(k+3)}\left[k\left(k^2+6 k+9\right)+4\right]$
$\begin{aligned}
&=\frac{k^3+6 k^2+9 k+4}{4(k+1)(k+2)(k+3)} \\
&=\frac{(k+1)\left(k^2+5 k+4\right)}{4(k+1)(k+2)(k+3)} \\
&=\frac{(k+1)(k+4)}{4(k+2)(k+3)}=\frac{(\overline{k+1})(\overline{k+1}+3)}{4(\overline{k+1})(\overline{k+1}+2)}
\end{aligned}$
This shows $P(n)$ is true for $n=k+1$ i.e., $P(k+1)$ is true whenever $P(k)$ is true Hence $P(n)$ is true for all $n \in N$