Let $P(n): 1+\frac{1}{(1+2)}+\frac{1}{(1+2+3)}+\ldots \ldots \ldots \ldots$
$+\frac{1}{(1+2+3+\ldots \ldots+n)}=\frac{2 n}{n+1} \quad \ldots(i)$
putting n=1,
$\mathrm{LHS}=1$, and $\mathrm{RHS}=\frac{2.1}{1+1}=\frac{2}{2}=1$
LHS $=$ RHS $\quad \therefore P(1)$ is true
Let $P(k)$ be true.
Putting $n=k$ then
$P(k): 1+\frac{1}{(1+2)}+\ldots \ldots \ldots .+\frac{1}{(1+2+3+\ldots \ldots+k)}=\frac{2 k}{k+1} \quad \ldots(i)$
Now we shall prove that $P(k+1)$ is true whenever $P(k)$ is true.
$P(k+1)$ :
$\begin{aligned}
&1+\frac{1}{(1+2)}+\ldots \ldots \ldots+\frac{1}{(1+2+3+\ldots . .+k)} \\
&+\frac{1}{(1+2+3+\ldots .(k+1))}
\end{aligned}$
$\begin{aligned}
&=\frac{2 k}{k+1}+\frac{1}{1+2+3+\ldots(k+1)}[\text { from(ii) }] \\
&=\frac{2 k}{k+1}+\frac{1}{\frac{(k+1)(k+2)}{2}} \\
&=\frac{2 k(k+2)+2}{(k+1)(k+2)}=\frac{2[k(k+2)+1]}{(k+1)(k+2)} \\
&=\frac{2\left(k^2+2 k+1\right)}{(k+1)(k+2)}=\frac{2(k+1)^2}{(k+1)(k+2)}=\frac{2(k+1)}{k+2}
\end{aligned}$
$\therefore P(k+1)$ is true whenever $P(k)$ is true.
Hence, $P(k)$ is true for all $n \in N$