## Precalculus (6th Edition) Blitzer

We use mathematical induction as follows: Statement ${{S}_{1}}$ is; ${{3}^{n-1}}=3$ And simplifying on the right, we obtain $3=\frac{{{3}^{n}}-1}{2}$. 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, ${{S}_{k}}=1+3+{{3}^{2}}+\ldots +{{3}^{k-1}}=\frac{{{3}^{k}}-1}{2}$ By adding ${{3}^{\left( k+1 \right)-1}}$ on both sides as: \begin{align} & 1+3+{{3}^{2}}+\ldots +{{3}^{k-1}}+{{3}^{\left( k+1 \right)-1}}=\frac{{{3}^{k}}-1}{2}+{{3}^{\left( k+1 \right)-1}} \\ & 1+3+{{3}^{2}}+\ldots +{{3}^{k-1}}+{{3}^{\left( k+1 \right)-1}}=\frac{{{3}^{k}}-1}{2}+{{3}^{k}} \\ & 1+3+{{3}^{2}}+\ldots +{{3}^{k-1}}+{{3}^{\left( k+1 \right)-1}}=\frac{{{3}^{k}}-1+2\times {{3}^{k}}}{2} \\ & 1+3+{{3}^{2}}+\ldots +{{3}^{k-1}}+{{3}^{\left( k+1 \right)-1}}=\frac{\left( 2+1 \right){{3}^{k}}-1}{2} \end{align} And, \begin{align} & 1+3+{{3}^{2}}+\ldots +{{3}^{k-1}}+{{3}^{\left( k+1 \right)-1}}=\frac{\left( 3 \right){{3}^{k}}-1}{2} \\ & 1+3+{{3}^{2}}+\ldots +{{3}^{k-1}}+{{3}^{\left( k+1 \right)-1}}=\frac{{{3}^{k+1}}-1}{2} \end{align} Thus, ${{S}_{k+1}}$ is true. The result ${{S}_{n}}=1+3+{{3}^{2}}+....+{{3}^{n-1}}=\frac{{{3}^{n}}-1}{2}$ holds true.