University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Section 5.2 - Sigma Notation and Limits of Finite Sums - Exercises: 23

Answer

$-73$

Work Step by Step

$\sum\limits_{k=1}^6(3-k^2)=\sum\limits_{k=1}^63-\sum\limits_{k=1}^6k^2=6\cdot3-(1^2+2^2+3^2+4^2+5^2+6^2)=18-(1+4+9+16+25+36)=18-91=-73$
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