Answer
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 $r=3^2$.