Answer
Prove $n+ \frac{n}{2} + \frac{n}{4} + \cdots + \frac{n}{2^n}$ is $\Theta(n)$.
By theorem 5.2.3:
$n+ \frac{n}{2} + \frac{n}{4} + \cdots + \frac{n}{2^n} = n(1+ \frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2^n}) = n(\frac{(\frac{1}{2})^{n+1}-1}{\frac{1}{2}-1})
= n(\frac{1-2^{n+1}}{-2^n})$ (multiply numerator and denominator by $2^{n+1}$)
$=n(\frac{2^{n+1}-1}{2^n}) = n(2-\frac{1}{2^n})$.
For $n>1$,
$1 \leq (2-\frac{1}{2^n}) \leq 2$
so $1(n) \leq n(2-\frac{1}{2^n}) \leq 2(n)$.
By the transitive property:
$1(n) \leq n+ \frac{n}{2} + \frac{n}{4} + \cdots + \frac{n}{2^n} \leq 2(n)$.
Because all terms are positive:
$1|(n)| \leq |n+ \frac{n}{2} + \frac{n}{4} + \cdots + \frac{n}{2^n}| \leq 2|(n)|$.
Thus for A=1, B=2, k=1,
$A|(n)| \leq |n+ \frac{n}{2} + \frac{n}{4} + \cdots + \frac{n}{2^n}| \leq B|(n)|$ for all $x>k$.
Thus $n+ \frac{n}{2} + \frac{n}{4} + \cdots + \frac{n}{2^n}$ is $\Theta(n)$.
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 $n \geq 0$,
$\sum_{i=0}^{n}r^i = \frac{r^{n+1}-1}{r-1}$.
In this case $r=\frac{1}{2}$.