Answer
See the explanation below.
Work Step by Step
We use mathematical induction as follows:
Statement ${{S}_{1}}$ is;
${{3}^{n-1}}=3$
And simplifying on the right, we obtain $3=\frac{{{3}^{n}}-1}{2}$.
So, this true statement shows that ${{S}_{1}}$ is true.
And assume ${{S}_{k}}$ is true. Using ${{S}_{k}},{{S}_{k+1}}$ from the expression,
${{S}_{k}}=1+3+{{3}^{2}}+\ldots +{{3}^{k-1}}=\frac{{{3}^{k}}-1}{2}$
By adding ${{3}^{\left( k+1 \right)-1}}$ on both sides as:
$\begin{align}
& 1+3+{{3}^{2}}+\ldots +{{3}^{k-1}}+{{3}^{\left( k+1 \right)-1}}=\frac{{{3}^{k}}-1}{2}+{{3}^{\left( k+1 \right)-1}} \\
& 1+3+{{3}^{2}}+\ldots +{{3}^{k-1}}+{{3}^{\left( k+1 \right)-1}}=\frac{{{3}^{k}}-1}{2}+{{3}^{k}} \\
& 1+3+{{3}^{2}}+\ldots +{{3}^{k-1}}+{{3}^{\left( k+1 \right)-1}}=\frac{{{3}^{k}}-1+2\times {{3}^{k}}}{2} \\
& 1+3+{{3}^{2}}+\ldots +{{3}^{k-1}}+{{3}^{\left( k+1 \right)-1}}=\frac{\left( 2+1 \right){{3}^{k}}-1}{2}
\end{align}$
And,
$\begin{align}
& 1+3+{{3}^{2}}+\ldots +{{3}^{k-1}}+{{3}^{\left( k+1 \right)-1}}=\frac{\left( 3 \right){{3}^{k}}-1}{2} \\
& 1+3+{{3}^{2}}+\ldots +{{3}^{k-1}}+{{3}^{\left( k+1 \right)-1}}=\frac{{{3}^{k+1}}-1}{2}
\end{align}$
Thus, ${{S}_{k+1}}$ is true. The result ${{S}_{n}}=1+3+{{3}^{2}}+....+{{3}^{n-1}}=\frac{{{3}^{n}}-1}{2}$ holds true.
