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?


It's equivalent to

$a_{n} + 2 = 2\left(a_{n - 1} + 2\right) = 2^{2}\left(a_{n - 2} + 2\right) = \cdots =
2^{n}\left(a_{0} + 2\right) = 4 \times 2^{n}\implies
\bbox[10px,border:1px dotted navy]{\displaystyle a_{n} = 2\left(\,2^{n + 1} - 1\right)}
$
