Answer
$61$
Work Step by Step
$\sum\limits_{k=1}^6(k^2-5)=\sum\limits_{k=1}^6k^2-\sum\limits_{k=1}^65-=(1^2+2^2+3^2+4^2+5^2+6^2)-6\cdot5=(1+4+9+16+25+36)-30=91-30=61$
University Calculus: Early Transcendentals (3rd Edition)
by Hass, Joel R.; Weir, Maurice D.; Thomas Jr., George B.
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 - Page 299: 24
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