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 - Page 1048: 4


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

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 $\frac{1}{81}$.
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