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 - Chapter Test - Page 840: 4



Work Step by Step

$\sum_{k=1}^{4} [(\frac{2}{3})^k-k]\\=[(\frac{2}{3})^1-1]+[(\frac{2}{3})^2-2]+[(\frac{2}{3})^3-3]+[(\frac{2}{3})^4-4]\\=\frac{2}{3}+\frac{4}{9}+\frac{8}{27}+\frac{16}{81}-10\\=-\frac{680}{81}$
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