Answer
Prove $4+4^2+4^3+ \cdots + 4^n$ is $\Theta(4^n)$.
By theorem 5.2.4: $4+4^2+4^3+ \cdots + 4^n = 4(1+4+4^2+4^3+ \cdots + 4^{n-2}) = 4(\frac{4^n-1}{4-1}) = \frac{4}{3}(4^n-1)$.
For $n>1$, $\frac{4}{3}(4^n-1)\leq \frac{4}{3}(4^{n+1}) = \frac{4^2}{3}(4^n)$.
For $n>1$, $4^n \leq \frac{4}{3}(4^n-1)$.
Since all terms are positive:
$|4^n| \leq |4+4^2+4^3+ \cdots + 4^n| \leq \frac{4^2}{3}|(4^n)|$.
Hence for A=1, B=$\frac{4^2}{3}$, k=1,
$A|4^n| \leq |4+4^2+4^3+ \cdots + 4^n| \leq B|(4^n)|$ for all $n>k$.
Hence $4+4^2+4^3+ \cdots + 4^n$ is $\Theta(4^n)$.
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 $n\geq 0$,
$\sum_{i=0}^{n}r^i = \frac{r^{n+1}-1}{r-1}$.