Precalculus (6th Edition) Blitzer

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

Chapter 10 - Review Exercises - Page 1125: 61


See the explanation below.

Work Step by Step

${{S}_{1}}$: 2 is a factor of ${{1}^{2}}+5\left( 1 \right)=6$ since $6=2\cdot 3$ ${{S}_{k}}$: 2 is a factor of ${{k}^{2}}+5k $ ${{S}_{k+1}}$: 2 is a factor of ${{\left( k+1 \right)}^{2}}+5\left( k+1 \right)$ $\begin{align} & {{\left( k+1 \right)}^{2}}+5\left( k+1 \right)={{k}^{2}}+2k+1+5k+5 \\ & ={{k}^{2}}+7k+6 \\ & ={{k}^{2}}+5k+2\left( k+3 \right) \\ & =\left( {{k}^{2}}+5k \right)+2\left( k+3 \right) \end{align}$ Since, we assume ${{S}_{k}}$ is true, then, we know 2 is a factor of ${{k}^{2}}+5k $. Since 2 is a factor of $2\left( k+3 \right)$, we conclude 2 is a factor of the sum $\left( {{k}^{2}}+5k \right)+2\left( k+3 \right)$. If ${{S}_{k}}$ is true, then ${{S}_{k+1}}$ is also true.
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