Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 11 - Section 11.3 - Series - Exercise Set - Page 651: 25

Answer

$\displaystyle\sum_{i=1}^{4}\left[12\left( \dfrac{1}{3} \right)^{i-1}\right]$

Work Step by Step

Using the summation notation, the series $ 12+4+\dfrac{4}{3}+\dfrac{4}{9} $ is equivalent to \begin{array}{l} \displaystyle\sum_{i=1}^{4}\left[12\left( \dfrac{1}{3} \right)^{i-1}\right] .\end{array}
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