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: 16

Answer

$\sum\limits_{k=1}^5(-1)^{k}\cdot\frac{k}{5}$

Work Step by Step

The denominators of the fractions are fives, the sign is always changing, and the numerator always grows by one, therefore the formula is: $\sum\limits_{k=1}^5(-1)^{k}\cdot\frac{k}{5}$
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