Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 12 - Sequences; Induction; the Binomial Theorem - Chapter Review - Review Exercises - Page 838: 5

Answer

$\sum_{k=1}^{4} (4k+2)=6+10+14+18=48$

Work Step by Step

Expand by substituing $k=1$ to $k=4$ to obtain: $\sum_{k=1}^{4} (4k+2)=(4(1)+2)+(4(2)+2)+(4(3)+2)+(4(4)+2)=6+10+14+18=48$
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