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: 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: $\sum\limits_{k=1}^5(-1)^{k}\cdot\frac{k}{5}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.