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: 29

Answer

$a_{1}=(-\frac{1}{2})^{1-1}=(-\frac{1}{2})^{0}=1$ $a_{2}=(-\frac{1}{2})^{2-1}=(-\frac{1}{2})^{1}=-\frac{1}{2}$ $a_{3}=(-\frac{1}{2})^{3-1}=(-\frac{1}{2})^{2}=\frac{1}{4}$ $a_{4}=(-\frac{1}{2})^{4-1}=(-\frac{1}{2})^{3}=-\frac{1}{8}$ $a_{10}=(-\frac{1}{2})^{10-1}=(-\frac{1}{2})^{9}=-\frac{1}{512}$ $a_{15}=(-\frac{1}{2})^{15-1}=(-\frac{1}{2})^{14}=\frac{1}{16384}$

Work Step by Step

If we want to find a term, we have to substitute $n$ by its index: $a_{1}=(-\frac{1}{2})^{1-1}=(-\frac{1}{2})^{0}=1$ $a_{2}=(-\frac{1}{2})^{2-1}=(-\frac{1}{2})^{1}=-\frac{1}{2}$ $a_{3}=(-\frac{1}{2})^{3-1}=(-\frac{1}{2})^{2}=\frac{1}{4}$ $a_{4}=(-\frac{1}{2})^{4-1}=(-\frac{1}{2})^{3}=-\frac{1}{8}$ $a_{10}=(-\frac{1}{2})^{10-1}=(-\frac{1}{2})^{9}=-\frac{1}{512}$ $a_{15}=(-\frac{1}{2})^{15-1}=(-\frac{1}{2})^{14}=\frac{1}{16384}$
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.