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: 5

Answer

$\frac{\sqrt 3-2}{2}$

Work Step by Step

$\sum\limits_{k=1}^3(-1)^{k+1}sin \frac{\pi}{k}=(-1)^{1+1}sin \frac{\pi}{1}+(-1)^{2+1}sin \frac{\pi}{2}+(-1)^{3+1}sin \frac{\pi}{3}=(-1)^{2}sin \frac{\pi}{1}+(-1)^{3}sin \frac{\pi}{2}+(-1)^{4}sin \frac{\pi}{3}=sin \frac{\pi}{1}+-sin \frac{\pi}{2}+sin \frac{\pi}{3}=0-1+\frac{\sqrt 3}{2}=\frac{\sqrt 3-2}{2}$
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