why$2 (-1 + 2^{1 + n})$ is the answer to the recurrence relation $a_{n}=2a_{n-1}+2$? $a_{0}=2$

$a_{1}=2(2)+2$

$a_{2}=2(2(2)+2)+2$

$a_{3}=2(2(2(2)+2)+2)+2$

$a_{4}=2(2(2(2(2)+2)+2)+2)+2$

$a_{5}=2(2(2(2(2(2)+2)+2)+2)+2)+2$

To simplifiy

$a_{6}=2^{6}+2^{5}...2^{1}$

so my answer is 

$a_{n}=2^{n+1}+2^{n}+...2^{1}$

The correct answer is $2 (-1 + 2^{1 + n})$

How do I make this transition?
Your expression for $a_n$ is correct. So $$a_n=2+\cdots+2^n+2^{n+1}.$$ Now note that this is a sum of $n+1$ terms in geometric progression with first term $a=2$ and the common ratio $r=2$. The sum $S$ can be found by $$S=\frac{a(r^{n+1}-1)}{r-1}$$ So substituting $a=2$ and $r=2$ we get $$a_n=\frac{2(2^{n+1}-1)}{2-1}=2(2^{n+1}-1)$$ as required.
You can also do this in base 2 fairly easily:

$$
\begin{align*}
a_0 &amp;= 10_2 \\
a_1 &amp;= 2a_0 + 2 = 100_2 + 10_2 = 110_2 \\
a_2 &amp;= 1100_2 + 10_2 = 1110_2.
\end{align*}
$$

We can see that by induction we will have.
$$a_n = \underbrace{11\cdots1}_{n+1}0_2 = 2(\underbrace{11\cdots1_2}_{n+1}).$$
Now note that
$$\underbrace{11\cdots1_2}_{n+1} + 1 = 1\underbrace{00\cdots0_2}_{n+1} = 2^{n + 1}.$$
Therefore
$$a_n = 2(2^{n + 1} - 1).$$