1 + 3 + 3

Let P(n) be the given statement.

Now,

$P(n)=1+3+3^{2}+\ldots+3^{n-1}=\frac{3^{n}-1}{2}$

Step 1:

$P(1)=1=\frac{3^{1}-1}{2}=\frac{2}{2}=1$

Hence, $P(1)$ is true.

Step 2:

Let $P(m)$ is true.

then,

$1+3+3^{2}+\ldots+3^{m-1}=\frac{3^{m}-1}{2}$

We shall prove that $P(m+1)$ is true.

That is,

$1+3+3^{2}+\ldots+3^{m}=\frac{3^{m+1}-1}{2}$

Now, we have:

$1+3+3^{2}+\ldots+3^{m-1}=\frac{3^{m}-1}{2}$

$\Rightarrow 1+3+3^{2}+\ldots+3^{m-1}+3^{m}=\frac{3^{m}-1}{2}+3^{m} \quad$ [Adding $3^{m}$ to both sides]

$\Rightarrow 1+3+3^{2}+\ldots+3^{m}=\frac{3^{m}-1+2 \times 3^{m}}{2}=\frac{3^{m}(1+2)-1}{2}=\frac{3^{m+1}-1}{2}$

Hence, $P(m+1)$ is true.

By the $p$ rinciple of $m$ athematical $i$ nduction, $P(n)$ is true for all $\mathrm{n} \in \mathrm{N}$.