Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 11 - Further Topics in Algebra - 11.1 Sequences and Series - 11.1 Exercises - Page 1014: 16

Answer

$$\frac{1}{2},\,\frac{1}{2},\,\frac{3}{8},\,\frac{1}{4},\,\frac{5}{{32}}$$

Work Step by Step

$$\eqalign{ & {a_n} = {\left( {\frac{1}{2}} \right)^n}\left( n \right) \cr & \cr & {\text{Complete for }}n = 1,2,3,4,5 \cr & n = 1 \cr & {a_1} = {\left( {\frac{1}{2}} \right)^1}\left( 1 \right) = \frac{1}{2} \cr & \cr & n = 2 \cr & {a_2} = {\left( {\frac{1}{2}} \right)^2}\left( 2 \right) = \frac{1}{2} \cr & \cr & n = 3 \cr & {a_3} = {\left( {\frac{1}{2}} \right)^3}\left( 3 \right) = \frac{3}{8} \cr & \cr & n = 4 \cr & {a_4} = {\left( {\frac{1}{2}} \right)^4}\left( 4 \right) = \frac{1}{4} \cr & \cr & n = 5 \cr & {a_5} = {\left( {\frac{1}{2}} \right)^5}\left( 5 \right) = \frac{5}{{32}} \cr & \cr & {\text{The first five terms of the sequence are:}} \cr & \frac{1}{2},\,\frac{1}{2},\,\frac{3}{8},\,\frac{1}{4},\,\frac{5}{{32}} \cr} $$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.