Precalculus (6th Edition)

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

Chapter 11 - Further Topics in Algebra - Summary Exercises on Sequences and Series - Exercises - Page 1038: 18

Answer

$\frac{1111}{500}$

Work Step by Step

1. We can identify the sequence as geometrical with $a_1=2, r=\frac{1}{10}$ 2. We have $\sum_{j=1}^42(\frac{1}{10})^{j-1}=2\times\frac{1-(\frac{1}{10})^4}{1-(\frac{1}{10})}=\frac{1111}{500}$
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