Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 10 - Section 10.1 - Sequences and Summation Notation - Exercise Set: 6

Answer

-$\frac{1}{3}$, $\frac{1}{9}$, -$\frac{1}{27}$ , $\frac{1}{81}$

Work Step by Step

To find the first four terms of the sequence whose general term is $a_{n}$ = $(\frac{1}{-3})^{n}$, we replace n in the formula with 1,2,3, and 4. n=1, $a_{1}$ = $(\frac{1}{-3})^{1}$ = -$\frac{1}{3}$ n=2, $a_{2}$ =$(\frac{1}{-3})^{2}$ = $\frac{1}{-3}$* $\frac{1}{-3}$ = $\frac{1}{9}$ n=3, $a_{3}$ = $(\frac{1}{-3})^{3}$ = $\frac{1}{-3}$*$\frac{1}{-3}$* $\frac{1}{-3}$ = -$\frac{1}{27}$ n=4,$a_{4}$ =$(\frac{1}{-3})^{4}$ = $\frac{1}{-3}$*$\frac{1}{-3}$* $\frac{1}{-3}$* $\frac{1}{-3}$ = $\frac{1}{81}$ The first four terms are -$\frac{1}{3}$, $\frac{1}{9}$, -$\frac{1}{27}$ and $\frac{1}{81}$.
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.