## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 10 - Section 10.4 - Mathematical Induction - Exercise Set - Page 1085: 19

#### 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.

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