## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 10 - Section 10.1 - Sequences and Summation Notation - Exercise Set - Page 1048: 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 $\frac{1}{81}$.

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