Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 10 - Section 10.4 - Mathematical Induction - Exercise Set - Page 1085: 25


See the explanation below.

Work Step by Step

We use mathematical induction as follows: Statement ${{S}_{1}}$ is: $\begin{align} & {{n}^{2}}-n={{1}^{2}}-1 \\ & =0 \end{align}$ $2$ is a factor of $0$ So, this statement shows that ${{S}_{1}}$ is true. Also, assume ${{S}_{k}}$ is true. Using ${{S}_{k}},{{S}_{k+1}}$ from the expression, ${{S}_{k}}:2$ is a factor of ${{k}^{2}}-k $ To show, ${{S}_{k+1}}:2$ is a factor of ${{\left( k+1 \right)}^{2}}-\left( k+1 \right)$ Consider, $\begin{align} & {{\left( k+1 \right)}^{2}}-\left( k+1 \right)={{k}^{2}}+2k+1-k-1 \\ & ={{k}^{2}}+k \\ & =\left( {{k}^{2}}-k \right)+2k \end{align}$ Since, 2 is a factor of ${{k}^{2}}-k $ and 2k, hence it is a factor of $\left( {{k}^{2}}-k \right)+2k $ So, 2 is a factor of ${{\left( k+1 \right)}^{2}}-\left( k+1 \right)$. Therefore, ${{S}_{k+1}}$ is true. The result ${{S}_{n}}:2$ is a factor of ${{n}^{2}}-n $ is true by mathematical induction. Thus, the provided statement is proved by mathematical induction.
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