Answer
See the explanation below.
Work Step by Step
We use mathematical induction as follows:
Statement ${{S}_{1}}$ is:
${{2}^{n}}=2$
And simplifying on the right, we obtain ${{2}^{n+1}}-2=2$.
So, this true statement shows that ${{S}_{1}}$ is true.
Also, assume ${{S}_{k}}$ is true. Using ${{S}_{k}},{{S}_{k+1}}$ from the expression,
${{S}_{k}}=2+4+8+\ldots +{{2}^{k}}={{2}^{k+1}}-2$
By adding ${{2}^{k+1}}$ on both sides, we get:
$\begin{align}
& 2+4+8+\ldots +{{2}^{k}}+{{2}^{k+1}}={{2}^{k+1}}-2+{{2}^{k+1}} \\
& 2+4+8+\ldots +{{2}^{k}}+{{2}^{k+1}}=2\times {{2}^{k+1}}-2 \\
& 2+4+8+\ldots +{{2}^{k}}+{{2}^{k+1}}={{2}^{k+2}}-2 \\
\end{align}$
Thus, ${{S}_{k+1}}$ is true. The result ${{S}_{n}}=2+4+8+\ldots +{{2}^{n}}={{2}^{n+1}}-2$ holds true.