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: 12

Answer

$\sum\limits_{k=1}^4k^2$

Work Step by Step

$1=1^2, 4=2^2, 9=3^2, 16=4^2$, therefore the additives are the first four square numbers, therefore the formula is: $\sum\limits_{k=1}^4k^2$
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