College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 8 - Sequences, Induction, and Probability - Exercise Set 8.4 - Page 751: 8


See below.

Work Step by Step

$S_k:2+7+12+...+(5k-3)=\frac{k(5k-1)}{2}$ $S_{k+1}:2+7+12+...+(5(k+1)-3)=\frac{(k+1)(5(k+1)-1)}{2}\\S_{k+1}:2+7+12+...+(5k+2)=\frac{(k+1)(5k+4)}{2}$
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