Discrete Mathematics with Applications 4th Edition

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

Chapter 4 - Elementary Number Theory and Methods of Proof - Exercise Set 4.2 - Page 169: 27

Answer

Yes. $\frac{2^{n+1}-1}{2^{n}}$

Work Step by Step

Yes. Simplify the right hand side as: $RHS=\frac{1-\frac{1}{2^{n+1}}}{1-\frac{1}{2}}=\frac{2^{n+1}-1}{2^{n+1}-2^{n}}=\frac{2^{n+1}-1}{2^{n}}$ As $n$ is a non-negative integer, we have $2^{n+1}-1$ and $2^n\ne0$ as integers, thus $RHS$ is a rational number.
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