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: 6

Answer

See below.

Work Step by Step

$S_k:3+4+5+...+(k+2)=\frac{k(k+5)}{2}$ $S_{k+1}:3+4+5+...+((k+1)+2)=\frac{(k+1)((k+1)+5)}{2}\\S_{k+1}:3+4+5+...+((k+3)=\frac{(k+1)(k+6)}{2}$
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