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: 29

Answer

$\frac{1-(-2)^{n+1}}{3}$

Work Step by Step

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