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 - Page 299: 14


$\sum\limits_{k=1}^5 2\cdot k$

Work Step by Step

$2=2\cdot1, 4=2\cdot2, 6=2\cdot3, 8=2\cdot4, 10=2\cdot5$, therefore the kth element is $2\cdot k$, therefore the formula is: $\sum\limits_{k=1}^5 2\cdot k$
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