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?

The inhomogeneous recurrence relation
$$
a_n = 2 a_{n-1} + 2
$$
can be turned into a homogeneous recurrence
$$
a_n - a_{n-1} = 2 a_{n-1} + 2 - (2 a_{n-2} + 2) = 2 a_{n-1} - 2 a_{n-2} \iff \\
a_n = 3 a_{n-1} - 2 a_{n-2}
$$
and solved by the usual algorithm.
The characteristic polynomial is
$$
p(t) = t^2 - 3 t + 2
$$
with roots $r_1 = 1$ and $r_2 = 2$.
So the solution is
$$
a_n = k_1 r_1^n + k_2 r_2^n = k_1 + k_2 2^n
$$
The initial elements give
$$
a_0 = k_1 + k_2 = 2 \\
a_1 = k_1 + 2 k_2 = 6
$$
This gives $k_2 = 4$ and $k_1 = -2$. The solution is
$$
a_n = -2 + 4 \cdot 2^n = 2^{n+2} - 2
$$
