Answer
See the explanation below.
Work Step by Step
We use mathematical induction as follows:
Statement ${{S}_{1}}$ is;
$\frac{1}{2}=\frac{1}{{{2}^{n}}}$
And simplifying on the right, we obtain $\frac{1}{2}=1-\frac{1}{{{2}^{n}}}$.
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,
${{S}_{k}}=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{{{2}^{k}}}=1-\frac{1}{{{2}^{k}}}$
By adding $\frac{1}{{{2}^{k+1}}}$ on both sides, we get:
$\begin{align}
& \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{{{2}^{k}}}+\frac{1}{{{2}^{k+1}}}=1-\frac{1}{{{2}^{k}}}+\frac{1}{{{2}^{k+1}}} \\
& \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{{{2}^{k}}}+\frac{1}{{{2}^{k+1}}}=1+\frac{{{2}^{k}}-{{2}^{k+1}}}{{{2}^{k+1}}{{2}^{k}}} \\
& \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{{{2}^{k}}}+\frac{1}{{{2}^{k+1}}}=1+\frac{{{2}^{k}}-{{2}^{k+1}}}{{{2}^{2k+1}}} \\
& \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{{{2}^{k}}}+\frac{1}{{{2}^{k+1}}}=1-\frac{1}{{{2}^{k+1}}} \\
\end{align}$
Thus, ${{S}_{k+1}}$ is true. The result $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{{{2}^{n}}}=1-\frac{1}{{{2}^{n}}}$ holds true.
