Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 11 - Section 11.3 - Series - Exercise Set - Page 651: 34

Answer

$-\dfrac{47}{120}$

Work Step by Step

Given that $ a_n=\dfrac{(-1)^n}{2n} $ then the first 5 terms are \begin{array}{l} a_1=\dfrac{(-1)^1}{2(1)}\\\\ a_1=\dfrac{-1}{2}\\\\ a_1=-\dfrac{1}{2} ,\\\\ a_2=\dfrac{(-1)^2}{2(2)}\\\\ a_2=\dfrac{1}{4} ,\\\\ a_3=\dfrac{(-1)^3}{2(3)}\\\\ a_3=\dfrac{-1}{6}\\\\ a_3=-\dfrac{1}{6} ,\\\\ a_4=\dfrac{(-1)^4}{2(4)}\\\\ a_4=\dfrac{1}{8} ,\\\\ a_5=\dfrac{(-1)^5}{2(5)}\\\\ a_5=\dfrac{-1}{10}\\\\ a_5=-\dfrac{1}{10} .\end{array} Hence, the sum is $ -\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{8}-\dfrac{1}{10} = -\dfrac{47}{120} .$
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