College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 9 - Section 9.1 - Sequences - 9.1 Assess Your Understanding - Page 647: 58



Work Step by Step

There are n-2 terms, as the index k changes from 3 to n. The index k dictates how the terms are formed: $\displaystyle \sum_{k=3}^{n}(-1)^{k+1}2^{k}=$ $=(-1)^{3+1}2^{3}+(-1)^{4+1}2^{4}+(-1)^{5+1}2^{5}+...+(-1)^{n+1}2^{n}$ $=(-1)^{4}2^{3}+(-1)^{5}2^{4}+(-1)^{6}2^{5}+\cdots+(-1)^{n+1}2^{n}$ When the exponent of $(-1)^{n}$ is odd, we have $-1$ and when it is even, $+1.$ $=2^{3}-2^{4}+2^{5}-2^{6}+\cdots+(-1)^{n+1}2^{n}$ $=8-16+32-64+\ldots+(-1)^{n+1}2^{n}$
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