Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 11 - Sequences; Induction; the Binomial Theorem - Section 11.1 Sequences - 11.1 Assess Your Understanding - Page 827: 58

Answer

$8-16+32-64+........+(-1)^{n+1}2^{n}$

Work Step by Step

We see that there are $n-2$ terms, as the index k changes from $3$ to $n$. The index $k$ indicates how the terms are formed. Thus, we write the sum for the $n$ term as follows: $\displaystyle \sum_{k=3}^{n}(-1)^{k+1}2^{k}=$ $=(-1)^{3+1}2^{3}+(-1)^{4+1}2^{4}+(-1)^{5+1}2^{5}+......+(-1)^{n+1}2^{n} \\=(-1)^{4}2^{3}+(-1)^{5}2^{4}+(-1)^{6}2^{5}+.....+(-1)^{n+1}2^{n}$ We find that we have alternating sign. The value of $(-1)^{n}$ is $-1$ when $n$ is odd and $+1$ when $n$ is even. $=2^{3}-2^{4}+2^{5}-2^{6}+.......+(-1)^{n+1}2^{n} \\ = 8-16+32-64+........+(-1)^{n+1}2^{n}$
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