Answer
$a_1=4$
$a_2=14$
$a_3=34$
$a_4=74$
$a_{5}=154$
Work Step by Step
We are given:
$a_{n}=2(a_{n-1}+3)$ and $a_1=4$
We evaluate:
$a_2=2(a_{1}+3)=2(4+3)=14$
$a_3=2(a_{2}+3)=2(14+3)=34$
$a_4=2(a_{3}+3)=2(34+3)=74$
$a_{5}=2(a_{4}+3)=2(74+3)=154$
College Algebra 7th Edition
by Stewart, James; Redlin, Lothar; Watson, Saleem
Published by
Brooks Cole
ISBN 10:
1305115546
ISBN 13:
978-1-30511-554-5
Chapter 8, Sequences and Series - Section 8.1 - Sequences and Summation Notation - 8.1 Exercises - Page 600: 15
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