Answer
See below.
Work Step by Step
$a_n=3\cdot2^{n-1}$ thus
$a_1=3\cdot2^{1-1}=3\cdot1=3$
$a_2=3\cdot2^{2-1}=3\cdot2=6$
$a_3=3\cdot2^{3-1}=3\cdot4=12$
$a_4=3\cdot2^{4-1}=3\cdot8=24$
$a_5=3\cdot2^{5-1}=3\cdot16=48$
College Algebra (6th Edition)
by Blitzer, Robert F.
Published by
Pearson
ISBN 10:
0-32178-228-3
ISBN 13:
978-0-32178-228-1
Chapter 8 - Summary, Review, and Test - Review Exercises - Page 789: 31
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