Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 12 Sequences and Series - 12.1 Define and Use Sequences and Series - 12.1 Exercises - Skill Practice - Page 799: 61


False. See counterexample below.

Work Step by Step

Let n be an even number, say 4. We will build a counterexample using $(-1)^{odd}=-1,\qquad (-1)^{even}=+1$ Let $a_{i}=b_{i}=(-1)^{i}$ Then, the sum on the LHS is $\displaystyle \sum_{i=1}^{4}a_{i}b_{i}=\sum_{i=1}^{4}(-1)^{i}(-1)^{i}$ $=\displaystyle \sum_{i=1}^{4}[(-1)(-1)]^{i}$ $=\displaystyle \sum_{i=1}^{4}1^{i}$ $=1+1+1+1=4$ but, $\displaystyle \sum_{i=1}^{4}a_{i}=(-1)^{1}+(-1)^{2}+(-1)^{3}+(-1)^{4}$ $=-1+1-1+1$ $=0$ The product on the RHS is $0.$ So, for this example, the LHS=4, and the RHS=0. The statement is not true.
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