Answer
$80$
Work Step by Step
We expand the sum:
$\displaystyle \sum_{i=4}^{8}(3i-2)$
$=[3(4)-2]+[3(5)-2]+[3(6)-2]+[3(7)-2]+[3(8)-2]$
$=(12-2)+(15-2)+(18-2)+(21-2)+(24-2)$
$=10+13+16+19+22$
$=80$
Calculus: Early Transcendentals 8th Edition
by Stewart, James
Published by
Cengage Learning
ISBN 10:
1285741552
ISBN 13:
978-1-28574-155-0
APPENDIX E - Sigma Notation - E Exercises - Page A 38: 21
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