Discrete Mathematics with Applications 4th Edition

Published by Cengage Learning
ISBN 10: 0-49539-132-8
ISBN 13: 978-0-49539-132-6

Chapter 5 - Sequences, Mathematical Induction, and Recursion - Exercise Set 5.1 - Page 243: 47

Answer

$$\sum_{k=1}^{6} (-1)^{k+1} \cdot r^{k-1}$$

Work Step by Step

All the terms involve $r$ so replace $1$ with its equivalent $r^0$ to obtain: $=r^0-r+r^2-r^3+r^4-r^5$ The terms alternate signs so the summation involves either $(-1)^k$ or $(-1)^{k+1}$. However, since the first term is positive, then the summation must involve $(-1)^{k+1}$ as this gives a positive value when $k=1$. The summation has 6 terms. This means that $k$ will vary from $k=1$ to $k=6$. The exponent of $r$ varies from $0$ to $5$. This means that when $k=1$, the exponent pf $r$ will be $k-1=0$. Thus, the exponent of $r$ is $k-1$. Therefore, the summation notation whose expanded form is given is: $$\sum_{k=1}^{6} (-1)^{k+1} \cdot r^{k-1}$$
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