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 - Concept and Vocabulary Check - Page 1084: 4

Answer

Now, consider the statement: $2$ is a factor of ${{n}^{2}}+3n $ If $ n=1$, the statement is $2$ is a factor of $1+3$ If $ n=2$, the statement is $2$ is a factor of $4+6$ If $ n=3$, the statement is $2$ is a factor of $9+9$ If $ n=k+1$, the statement before the algebra is simplified is $2$ is a factor of ${{\left( k+1 \right)}^{2}}+3\left( k+1 \right)$ If $ n=k+1$, the provided statement after the algebra is simplified is $2$ is a factor of ${{k}^{2}}+5k+4$

Work Step by Step

For the statement $2$ is a factor of ${{n}^{2}}+3n $: if $ n=1$ substituting, $\begin{align} & {{n}^{2}}+3n={{1}^{2}}+3\times 1 \\ & =1+3 \\ & =4 \end{align}$ If the value of $ n=2$: $\begin{align} & {{n}^{2}}+3n={{2}^{2}}+3\times 2 \\ & =4+6 \\ & =10 \end{align}$ Then, the value $ n=3$ gives $\begin{align} & {{n}^{2}}+3n={{3}^{2}}+3\times 3 \\ & =9+9 \\ & =18 \end{align}$ If $ n=k+1$, $2$ is a factor ${{\left( k+1 \right)}^{2}}+3\left( k+1 \right)$ Further simplification gives, $\begin{align} & {{\left( k+1 \right)}^{2}}+3\left( k+1 \right)={{k}^{2}}+2k+1+3k+3 \\ & ={{k}^{2}}+5k+4 \end{align}$
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