Answer
$-\frac{680}{81}$
Work Step by Step
$\sum_{k=1}^{4} [(\frac{2}{3})^k-k]\\=[(\frac{2}{3})^1-1]+[(\frac{2}{3})^2-2]+[(\frac{2}{3})^3-3]+[(\frac{2}{3})^4-4]\\=\frac{2}{3}+\frac{4}{9}+\frac{8}{27}+\frac{16}{81}-10\\=-\frac{680}{81}$
Precalculus (10th Edition)
by Sullivan, Michael
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
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