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 764: 37


Prove $2+2\cdot3^2 + 2\cdot3^4+ \cdots+ 2\cdot3^{2n}$ is $\Theta(3^{2n})$. By theorem 5.2.3: $2+2\cdot3^2 + 2\cdot3^4+ \cdots+ 2\cdot3^{2n} = 2(\frac{3^{2n+2}-1}{3^2-1}) = \frac{1}{4}(3^{2n+2}-1)$. For $n>0$,$\frac{1}{4}(3^{2n+2}-1) \leq 3^{2n+2} = 9(3^{2n})$. By the transitive property: $2+2\cdot3^2 + 2\cdot3^4+ \cdots+ 2\cdot3^{2n} \leq 9(3^{2n})$. For $n>0$, $2\cdot3^{2n} \leq 2+2\cdot3^2 + 2\cdot3^4+ \cdots+ 2\cdot3^{2n}$. Since all terms are positive: $2|(3^{2n})| \leq |2+2\cdot3^2 + 2\cdot3^4+ \cdots+ 2\cdot3^{2n}| \leq 9|(3^{2n})|$. Therefore for A=2, B=9, k=1, $A|(3^{2n})| \leq |2+2\cdot3^2 + 2\cdot3^4+ \cdots+ 2\cdot3^{2n}| \leq B|(3^{2n})|$ for all $x>k$. Thus $2+2\cdot3^2 + 2\cdot3^4+ \cdots+ 2\cdot3^{2n}$ is $\Theta(3^{2n})$.

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 $n \geq 0$, $\sum_{i=0}^{n}r^i = \frac{r^{n+1}-1}{r-1}$. In this case $r=3^2$.
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