Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 8 - Sequences and Infinite Series - 8.1 An Overview - 8.1 Exercises: 11

Answer

$\frac{-1}{2},\frac{1}{4},\frac{-1}{8},\frac{1}{16}$

Work Step by Step

$a_n = \frac{(-1)^n}{2^n}$ To find the first four terms, we need to find $a_1, a_2, a_3$, and $a_4$. $a_1 = \frac{(-1)^1}{2^1} = \frac{-1}{2}$ $a_2= \frac{(-1)^2}{2^2} = \frac{1}{4}$ $a_3 = \frac{(-1)^3}{2^3} = \frac{-1}{8}$ $a_4 = \frac{(-1)^4}{2^4} = \frac{1}{16}$
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