Answer
See the explanation below.
Work Step by Step
$\begin{align}
& {{S}_{1}}:1=\frac{{{4}^{1}}-1}{3} \\
& 1=\frac{1}{3} \\
& 1=1\text{ is true}\text{.} \\
& {{S}_{k}}:1+4+{{4}^{2}}+....+{{4}^{k-1}}=\frac{{{4}^{k}}-1}{3} \\
\end{align}$
${{S}_{k+1}}:1+4+{{4}^{2}}+....+{{4}^{k-1}}+{{4}^{k}}=\frac{{{4}^{k+1}}-1}{3}$
And add ${{4}^{k}}$ to both sides of ${{S}_{k}}$:
$\begin{align}
& {{S}_{k}}:1+4+{{4}^{2}}+....+{{4}^{k-1}}=\frac{{{4}^{k}}-1}{3} \\
& 1+4+{{4}^{2}}+....+{{4}^{k-1}}+{{4}^{k}}=\frac{{{4}^{k}}-1}{3}+{{4}^{k}} \\
\end{align}$
Then, simplify the right-hand side:
$\begin{align}
& \frac{{{4}^{k}}-1}{3}+{{4}^{k}}=\frac{{{4}^{k}}-1+3\cdot {{4}^{k}}}{3} \\
& \frac{{{4}^{k}}-1}{3}+{{4}^{k}}=\frac{4\cdot {{4}^{k}}-1}{3} \\
& \frac{{{4}^{k}}-1}{3}+{{4}^{k}}=\frac{{{4}^{k+1}}-1}{3} \\
\end{align}$
If ${{S}_{k}}$ is true, then ${{S}_{k+1}}$ is true.
