Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 12 Sequences and Series - 12.1 Define and Use Sequences and Series - 12.1 Exercises - Skill Practice - Page 798: 14

Answer

$1,\ \displaystyle \frac{2}{3},\ \displaystyle \frac{3}{5}$, $\displaystyle \frac{4}{7}$, $\displaystyle \frac{5}{9}$, $\displaystyle \frac{6}{11}$

Work Step by Step

$f(1)=\displaystyle \frac{1}{2(1)-1}=1$ $f(2)=\displaystyle \frac{2}{2(2)-1}=\frac{2}{3}$ $f(3)=\displaystyle \frac{3}{2(3)-1}=\frac{3}{5}$ $f(4)=\displaystyle \frac{4}{2(4)-1}=\frac{4}{7}$ $f(5)=\displaystyle \frac{5}{2(5)-1}=\frac{5}{9}$ $f(6)=\displaystyle \frac{6}{2(6)-1}=\frac{6}{11}$
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