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: 24

Answer

$61$

Work Step by Step

$\sum\limits_{k=1}^6(k^2-5)=\sum\limits_{k=1}^6k^2-\sum\limits_{k=1}^65-=(1^2+2^2+3^2+4^2+5^2+6^2)-6\cdot5=(1+4+9+16+25+36)-30=91-30=61$
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