## Precalculus (6th Edition) Blitzer

We use mathematical induction as follows: Statement ${{S}_{1}}$ is: ${{2}^{n-1}}=1$ And simplifying on the right, we obtain ${{2}^{n}}-1=1$. 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, \begin{align} & {{S}_{k}}=1+2+{{2}^{2}}+\ldots +{{2}^{k-1}} \\ & ={{2}^{k}}-1 \end{align} Adding ${{2}^{\left( k+1 \right)-1}}$ on both sides as: \begin{align} & 1+2+{{2}^{2}}+\ldots +{{2}^{\left( k+1 \right)-1}}={{2}^{k}}-1+{{2}^{\left( k+1 \right)-1}} \\ & 1+2+{{2}^{2}}+\ldots +{{2}^{\left( k+1 \right)-1}}={{2}^{k}}-1+{{2}^{\left( k \right)}} \\ & 1+2+{{2}^{2}}+\ldots +{{2}^{\left( k+1 \right)-1}}=2\times {{2}^{k}}-1 \\ & 1+2+{{2}^{2}}+\ldots +{{2}^{\left( k+1 \right)-1}}={{2}^{k+1}}-1 \end{align} Thus, ${{S}_{k+1}}$ is true. The result ${{S}_{n}}=1+2+{{2}^{2}}+\ldots +{{2}^{n-1}}={{2}^{n}}-1$ holds true.