Answer
$6,12,24,48$
Work Step by Step
$f(n)=3(2^n)$
$f(1)=3\times2^1=6$
$f(2)=3\times2^2=12$
$f(3)=3\times2^3=24$
$f(4)=3\times2^4=48$
Intermediate Algebra: Connecting Concepts through Application
by Clark, Mark; Anfinson, Cynthia
Published by
Brooks Cole
ISBN 10:
0-53449-636-9
ISBN 13:
978-0-53449-636-4
Chapter 9 - Conic Sections, Sequences, and Series - 9.3 Arithmetic Sequences - 9.3 Exercises - Page 727: 4
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