Precalculus (6th Edition) Blitzer

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.