Let $P(n)$ be the given statement.
i.e., $P(n): 1+3+3^2+\ldots \ldots \ldots+3^{n-1}=\frac{\left(3^n-1\right)}{2}$
Putting $n=1, P(1)=\frac{3-1}{2}=1$
$P(n)$ is true for $n=1$.
Assume that $P(k)$ is true.
i.e., $P(k): 1+3+3^2+\ldots \ldots \ldots+3^{k-1}$
$=\frac{\left(3^k-1\right)}{2} \quad \ldots(i)$
We shall prove that $P(k+1)$ is true whenever $P(k)$ is true.
$\begin{aligned}
&\therefore P(k+1): 1+3+3^2+\ldots \ldots \ldots .+3^{(k+1)-1} \\
&=\frac{3^{k+1}-1}{2} \\
&{\left[1+3+3^2+\ldots \ldots \ldots+3^{k-1}\right]+3^k=\frac{\left(3^k-1\right)}{2}+3^k[\text { using }(\mathrm{i})]} \\
&=\frac{3^k-1+2.3^k}{2}=\frac{(1+2) 3^k-1}{2}=\frac{3.3^k-1}{2}=\frac{3^{k+1}-1}{2}
\end{aligned}$
$\therefore P(k+1)$ is also true whenever $P(k)$ is true.
Hence, $P(n)$ is true for all $n \in N$