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: 5


The required solution is ${{S}_{k}}=2{{k}^{2}}+2,{{S}_{k+1}}=\left( 2{{k}^{2}}+6k+4 \right)$

Work Step by Step

We are using ${{S}_{k}},{{S}_{k+1}}$ from the expression ${{S}_{n}}:4+8+12+...+4n=2n\left( n+1 \right)$ ${{S}_{k}}$ is given by, $\begin{align} & 4+8+12+...+4k=2k\left( k+1 \right) \\ & =2{{k}^{2}}+2 \end{align}$ ${{S}_{k+1}}$ Is provided by: $\begin{align} & 4+8+12+...+4\left( k+1 \right)=2\left( k+1 \right)\left( k+1+1 \right) \\ & =2\left( k+1 \right)\left( k+2 \right) \\ & =2\left( {{k}^{2}}+3k+2 \right) \\ & =\left( 2{{k}^{2}}+6k+4 \right) \end{align}$
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