Answer
$9+11+13+15+17=65$.
Work Step by Step
We need to sum the first five terms of the sequence whose general term is $a_{n}=2n+7$. This yields
$\sum\limits^{5}_{n=1}(2n+7)=[2(1)+7]+[2(2)+7]+[2(3)+7]+[2(4)+7]+[2(5)+7]$
$=9+11+13+15+17=65$
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 - 9.5 Series - 9.5 Exercises - Page 746: 3
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