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.
