College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 8, Sequences and Series - Chapter 8 Review - Exercises: 2

Answer

$a_1=-2$ $a_2=2$ $a_3=-\frac{8}{3}$ $a_4=4$ $a_{10}=\frac{512}{5}$

Work Step by Step

We are given: $a_{n}=(-1)^{n} \frac{2^{n}}{n}$ We evaluate: $a_1=(-1)^{1} \frac{2^{1}}{1}=-1*2=-2$ $a_2=(-1)^{2} \frac{2^{2}}{2}=1*4/2=2$ $a_3=(-1)^{3} \frac{2^{3}}{3}=-1*8/3=-\frac{8}{3}$ $a_4=(-1)^{4} \frac{2^{4}}{4}=1*16/4=4$ $a_{10}=(-1)^{10} \frac{2^{10}}{10}=\frac{1024}{10}=\frac{512}{5}$
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