Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 14 - Sequences, Series, and the Binomial Theorem - 14.1 Sequences and Series - 14.1 Exercise Set: 31

Answer

$a_{1}=\frac{(-1)^{1}}{1}=-1$ $a_{2}=\frac{(-1)^{2}}{2}=\frac{1}{2}$ $a_{3}=\frac{(-1)^{3}}{3}=-\frac{1}{3}$ $a_{4}=\frac{(-1)^{4}}{4}=\frac{1}{4}$ $a_{10}=\frac{(-1)^{10}}{10}=\frac{1}{10}$ $a_{15}=\frac{(-1)^{15}}{15}=-\frac{1}{15}$

Work Step by Step

If we want to find a term, we have to substitute $n$ by its index: $a_{1}=\frac{(-1)^{1}}{1}=-1$ $a_{2}=\frac{(-1)^{2}}{2}=\frac{1}{2}$ $a_{3}=\frac{(-1)^{3}}{3}=-\frac{1}{3}$ $a_{4}=\frac{(-1)^{4}}{4}=\frac{1}{4}$ $a_{10}=\frac{(-1)^{10}}{10}=\frac{1}{10}$ $a_{15}=\frac{(-1)^{15}}{15}=-\frac{1}{15}$
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