## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 10 - Section 10.4 - Mathematical Induction - Exercise Set - Page 1085: 5

#### Answer

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}

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.