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?
 A: 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.
A: 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
$$
a_n = -2 + 4 \cdot 2^n = 2^{n+2} - 2
$$
A: 
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)}
$
A: If you divide both sides by $2^n$, you will get :
$$\forall k\ge1,\,\frac{a_k}{2^k}-\frac{a_{k-1}}{2^{k-1}}=\frac1{2^{k-1}}$$
Now, cancellation occurs when summing up ($k$ from $1$ to $n$) :
$$\frac{a_n}{2^n}-a_0=\sum_{k=1}^n\frac1{2^{k-1}}=\frac{1-\frac1{2^n}}{1-\frac12}=2-\frac1{2^{n-1}}$$
that is :
$$a_n=2^n\left(4-\frac1{2^{n-1}}\right)=2^{n+2}-2$$
and finally :

$$a_n=2\left(2^{n+1}-1\right)$$

A: 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
$$
2^{-n}a_n=a_0+2-2^{1-n}\\
\Downarrow\\
\begin{align}
a_n
&=2^na_0+2^{n+1}-2\\[9pt]
&=2^{n+2}-2
\end{align}
$$
A: You can also do this in base 2 fairly easily:
$$
\begin{align*}
a_0 &= 10_2 \\
a_1 &= 2a_0 + 2 = 100_2 + 10_2 = 110_2 \\
a_2 &= 1100_2 + 10_2 = 1110_2.
\end{align*}
$$
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
$$a_n = 2(2^{n + 1} - 1).$$
