Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

APPENDIX E - Sigma Notation - E Exercises: 21

Answer

$80$

Work Step by Step

We expand the sum: $\displaystyle \sum_{i=4}^{8}(3i-2)$ $=[3(4)-2]+[3(5)-2]+[3(6)-2]+[3(7)-2]+[3(8)-2]$ $=(12-2)+(15-2)+(18-2)+(21-2)+(24-2)$ $=10+13+16+19+22$ $=80$
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