College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 8, Sequences and Series - Chapter 8 Review - Exercises: 3

Answer

$\displaystyle a_1=0$ $\displaystyle a_2=\frac{1}{4}$ $\displaystyle a_3=0$ $\displaystyle a_4=\frac{1}{32}$ $\displaystyle a_{10}=\frac{1}{500}$

Work Step by Step

We are given: $\displaystyle a_{n}= \frac{(-1)^{n}+1}{n^{3}}$ We evaluate: $\displaystyle a_1= \frac{(-1)^{1}+1}{1^{3}}=\frac{-1+1}{1}=0$ $\displaystyle a_2= \frac{(-1)^{2}+1}{2^{3}}=\frac{2}{8}=\frac{1}{4}$ $\displaystyle a_3=\frac{(-1)^{3}+1}{3^{3}}=\frac{-1+1}{27}=0$ $\displaystyle a_4= \frac{(-1)^{4}+1}{4^{3}}=\frac{2}{64}=\frac{1}{32}$ $\displaystyle a_{10}= \frac{(-1)^{10}+1}{10^{3}}=\frac{1+1}{1000}=\frac{1}{500}$
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