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 5.2 - Page 257: 28

Answer

2 - $\frac{1}{2^n}$

Work Step by Step

We know that, $1 + r + r^2 + ··· + r^n = \frac{r^{n+1}− 1}{r − 1}$ ---------------------------------- $1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+ ... +\frac{1}{2^n}$ = $1+\frac{1}{2}+(\frac{1}{2})^2+(\frac{1}{2})^3+ ... +(\frac{1}{2})^n$ = $\frac{(\frac{1}{2})^{n+1}− 1}{\frac{1}{2} − 1}$ =$\frac{(\frac{1}{2})^{n+1}− 1}{-\frac{1}{2}}$ =$2\times(1-(\frac{1}{2})^{n+1})$ =2 - $\frac{1}{2^n}$
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