Answer
$\displaystyle \sum_{i=1}^{6}(-1)^{i+1}(3)^i$
Work Step by Step
$3-9+27-81+243-729=(-1)^{1+1}(3)^1+(-1)^{2+1}(3)^2+(-1)^{3+1}(3)^3+(-1)^{4+1}(3)^4+(-1)^{5+1}(3)^5+(-1)^{6+1}(3)^6=\displaystyle \sum_{i=1}^{6}(-1)^{i+1}(3)^i$
Algebra and Trigonometry 10th Edition
by Larson, Ron
Published by
Cengage Learning
ISBN 10:
9781337271172
ISBN 13:
978-1-33727-117-2
Chapter 11 - 11.1 - Sequences and Series - 11.1 Exercises - Page 778: 83
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