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: 9


$\frac{2}{5}$,$\frac{2}{3}$ , $\frac{6}{7}$,1

Work Step by Step

To find the first four terms of the sequence whose general term is $a_{n}$ = $\frac{2n}{n+4}$, we replace n in the formula with 1,2,3, and 4. n=1, $a_{1}$ = $\frac{2n}{n+4}$ =$\frac{2*1}{1+4}$ = $\frac{2}{5}$ n=2, $a_{2}$ =$\frac{2n}{n+4}$ = $\frac{2*2}{2+4}$ = $\frac{4}{6}$ = $\frac{2}{3}$ n=3, $a_{3}$ = $\frac{2n}{n+4}$ =$\frac{2*3}{3+4}$ = $\frac{6}{7}$ n=4,$a_{4}$ = $\frac{2n}{n+4}$ = $\frac{2*4}{4+4}$ = $\frac{8}{8}$ =1 The first four terms are $\frac{2}{5}$,$\frac{2}{3}$ , $\frac{6}{7}$,1.
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