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: 10

Answer

$$183$$

Work Step by Step

$$\eqalign{ & \sum\limits_{i = 1}^5 {3{{\left( { - 3} \right)}^{i - 1}}} \cr & {\text{Let }}{a_n} = 3{\left( { - 3} \right)^{i - 1}} \cr & {a_1} = 3{\left( { - 3} \right)^{1 - 1}} = 3 \cr & {a_2} = 3{\left( { - 3} \right)^{2 - 1}} = - 9 \cr & {a_3} = 3{\left( { - 3} \right)^{3 - 1}} = 27 \cr & {a_4} = 3{\left( { - 3} \right)^{4 - 1}} = - 81 \cr & {a_5} = 3{\left( { - 3} \right)^{5 - 1}} = 243 \cr & {\text{Then,}} \cr & \sum\limits_{i = 1}^5 {3{{\left( { - 3} \right)}^{i - 1}}} = 3 - 9 + 27 - 81 + 243 \cr & {\text{Simplifying}} \cr & \sum\limits_{i = 1}^5 {3{{\left( { - 3} \right)}^{i - 1}}} = 183 \cr} $$
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