Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 12 - Sequences and Series - 12.1 Geometric Sequences - 12.1 Exercises - Page 612: 33

Answer

$${S_9} = \frac{{262143}}{2}$$

Work Step by Step

$$\eqalign{ & \sum\limits_{i = 0}^8 {\frac{3}{2}{{\left( 4 \right)}^i}} \cr & {S_n}{\text{ we write using summation notation as}}:{\text{ }}\left( {{\text{see page 612}}} \right) \cr & {S_n} = \sum\limits_{i = 0}^{n - 1} {a{r^i}} \cr & {\text{comparing the given sumation }}\sum\limits_{i = 0}^8 {\frac{3}{2}{{\left( 4 \right)}^i}} {\text{ with }}\sum\limits_{i = 0}^{n - 1} {a{r^i}} {\text{ we obtain }} \cr & a = \frac{3}{2}{\text{ and }}r = 4 \cr & {\text{the summation is from }}i = 0{\text{ to }}n - 1 = 8,{\text{ so }}n = 9 \cr & {\text{using the formula }}{S_n} = \frac{{a\left( {{r^n} - 1} \right)}}{{r - 1}},{\text{ }}r \ne 1{\text{ gives}} \cr & {S_9} = \frac{{\left( {3/2} \right)\left( {{4^9} - 1} \right)}}{{4 - 1}} \cr & {\text{simplify}} \cr & {S_9} = \frac{{{4^9} - 1}}{2} \cr & {S_9} = \frac{{262143}}{2} \cr} $$
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