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: 11

Answer

1, -$\frac{1}{3}$ , $\frac{1}{7}$, $\frac{-1}{15}$

Work Step by Step

To find the first four terms of the sequence whose general term is $a_{n}$ = $\frac{(-1)^{n+1}}{2^{n} -1}$. we replace n in the formula with 1,2,3, and 4. n=1, $a_{1}$ = $\frac{(-1)^{1+1}}{2^{1} -1}$ =$\frac{1}{1}$ = 1 n=2, $a_{2}$ =$\frac{(-1)^{2+1}}{2^{2} -1}$ = $\frac{-1}{3}$ = -$\frac{1}{3}$ n=3, $a_{3}$ = $\frac{(-1)^{3+1}}{2^{3} -1}$ =$\frac{1}{8 -1}$ = $\frac{1}{7}$ n=4,$a_{4}$ = $\frac{(-1)^{4+1}}{2^{4} -1}$ = $\frac{-1}{16 - 1}$ = $\frac{-1}{15}$ The first four terms are 1, -$\frac{1}{3}$ , $\frac{1}{7}$, $\frac{-1}{15}$ .
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