Answer
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 $n \geq 0$,
$\sum_{i=0}^{n}r^i = \frac{r^{n+1}-1}{r-1}$.
