Let $P(n): 1.3+2.3^2+3.3^3+\ldots \ldots . .+n .3^n$
$=\frac{(2 n-1) 3^{n+1}+3}{4}$
putting $n=1, P(1): \mathrm{LHS}=1.3=3$
$\mathrm{RHS}=\frac{(2-1) 3^2+3}{4}=\frac{12}{4}=3=$ LHS
This shows $P(n)$ is true for $n=1$
Let $P(n)$ be true for $n=k$
i.e., $P(k): 1.3+2.3^2+3.3^3+\ldots \ldots . .+k .3^k$
$=\frac{(2 k-1) 3^{k+1}+3}{4} \quad \ldots(i)$
We have to prove that $P(k+1)$ is true whenever $P(k)$ is true.
$\begin{aligned}
&P(k+1): 1.3+2.3^2+\ldots \ldots+(k+1) 3^{k+1} \\
&=\left[1.3+2.3^2+3.3^3+\ldots \ldots+k \cdot 3^k\right]+(k+1) 3^{k+1} \\
&=\frac{(2 k-1) \cdot 3^{k+1}+3}{4}+(k+1) \cdot 3^{k+1}[\text { using }(\mathrm{i})] \\
&=\frac{(2 k-1) 3^{k+1}+3+4(k+1) 3^{k+1}}{4} \\
&=\frac{3^{k+1}[2 k-1+4(k+1)]+3}{4} \\
&=\frac{3^{k+1}(6 k+3)+3}{4}=\frac{(2 k+1) 3^{k+2}+3}{4} \\
&=\frac{[2(k+1)-1] 3^{(k+1)+1}+3}{4}
\end{aligned}$
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 value $n \in N$.