Answer
61
Work Step by Step
6$\Sigma _{k=1}$(k^2-5)
=6$\Sigma _{k=1} k^2$- $ 6\Sigma _{k=1}$ 5
=$\frac{6(6+1)(2(6)+1)}{6} - 5(6) $
=61
Thomas' Calculus 13th Edition
by Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0-32187-896-5
ISBN 13:
978-0-32187-896-0
Chapter 5: Integrals - Section 5.2 - Sigma Notation and Limits of Finite Sums - Exercises 5.2 - Page 265: 24
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