Discrete Mathematics with Applications 4th Edition

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

Chapter 11 - Analysis of Algorithm Efficiency - Exercise Set 11.4 - Page 763: 35


Prove $1+2+2^2+ 2^3 + \cdots +2^n$ is $\Theta(2^n)$. By theorem 5.2.3: for all integers $n \geq 0$, $1+2+2^2+ 2^3 + \cdots +2^n = \frac{2^{n+1}-1}{2-1} = 2^{n+1}-1$ $2^{n+1}-1 \leq 2^{n+1} = 2(2^n)$ For $n>0$, $2^n \leq 2^{n+1}-1$ Because all terms are positive: $|2^n| \leq |1+2+2^2+ 2^3 + \cdots +2^n| \leq 2|2^n|$. Thus for A=1, B=2, k=0, $A|2^n| \leq |1+2+2^2+ 2^3 + \cdots +2^n| \leq B|2^n|$ for all $x>k$. Thus $1+2+2^2+ 2^3 + \cdots +2^n$ is $\Theta(2^n)$.

Work Step by Step

Recall the definition of $\Theta$-notation: $f(x)$ is $\Theta(g(x))$ iff there exist positive real numbers A, B, k, such that $A|g(x)| \leq |f(x)| \leq B|g(x)|$ for all $x>k$. Recall theorem 5.2.3: Sum of a geometric sequence: For any real number $r$ except 1, and any integer $n \geq 0$, $\sum_{i=0}^{n}r^i = \frac{r^{n+1}-1}{r-1}$.
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