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?

Let $a_n=2^nb_n$. Then
$$
a_{n}=2a_{n-1}+2\\\Downarrow\\2^nb_n=2^nb_{n-1}+2\\\Downarrow\\b_n=b_{n-1}+2^{1-n}
$$
Therefore,
$$
\begin{align}
b_n
&=b_0+\sum_{k=0}^{n-1}2^{-k}\\
&=b_0+\frac{1-2^{-n}}{1-2^{-1}}\\[9pt]
&=b_0+2-2^{1-n}
\end{align}
$$
and
$$
2^{-n}a_n=a_0+2-2^{1-n}\\
\Downarrow\\
\begin{align}
a_n
&=2^na_0+2^{n+1}-2\\[9pt]
&=2^{n+2}-2
\end{align}
$$
