Discrete Mathematics with Applications 4th Edition

Published by Cengage Learning
ISBN 10: 0-49539-132-8
ISBN 13: 978-0-49539-132-6

Chapter 5 - Sequences, Mathematical Induction, and Recursion - Exercise Set 5.1 - Page 242: 2

Answer

Just substitute $b_j$ by the index, starting from 1, because the sequence is defined for $j\geq1$, substitute the values and calculate.

Work Step by Step

$J\geq$ 1: $\begin{split} b_j & = \frac{5 - j}{5 + j}\\ & \\ b_1 & = \frac{5 - 1}{5 + 1} = \frac{4}{6} = \frac{2}{3}\\ & \\ b_2 & = \frac{5 - 2}{5 + 2} = \frac{3}{7} \\ & \\ b_3 & = \frac{5 - 3}{5 + 3} = \frac{2}{8} = \frac{1}{4} \\ & \\ b_4 & = \frac{5 - 4}{5 + 4} = \frac{1}{9} \\ & \\ \end{split}$
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