Answer
Show $2^{n+1}$ is $\Theta(2^n)$.
For all n, $0 \leq 2^n$
$2^n \leq 2^n + 2^n = 2(2^n) = 2^{n+1}$ (add $2^n$ to both sides)
$|2^n| \leq |2^{n+1}|$ (because all terms are positive)
$2^n \leq 2(2^n)$
$2(2^n) = 2^{n+1} \leq 4(2^n)$ (multiply both sides by 2)
$| 2^{n+1}| \leq 4|(2^n)|$ (because all terms are positive)
Hence for A=1, B=4, k=0,
$A|2^n| \leq |2^{n+1}| \leq B|2^n|$ for all $x>k$.
Thus $2^{n+1}$ 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 $x>k$.
