Help with solving this Recurrence Relation I really need help with this question
 Would anyone please give a simple step-by-step on how to solve this Recurrence Relation??


$a_n = 2a_{n-1} - 2a_{n-2}$ where $a_0 = 1$ and $a_1 = 3$


It would really be great if someone could explain how to solve this

So far I made it into the characteristic equation: 
$r^2 - 2r + 2 = 0$ 

But not sure where to go or if I am on the right track


Thanks

A generatingfunctionological solution would be to define $A(z) = \sum_{n \ge 0} a_n z^n$, write the recurrence shifting indices as:
$$
a_{n + 2} = 2 a_{n + 1} - 2 a_n
$$
Multiply the recurrence by $z^n$, sum over $n \ge 0$ and recognize some sums:
$$
\frac{A(z) - a_0 - a_1 z}{z^2}
  = 2 \frac{A(z) - a_0}{z} - 2 A(z)
$$
Plug in $a_0 = 1, a_1 = 3$ and solve for $A(z)$ to get
$$
A(z) = \frac{1 + z}{1 - 2 z + 2 z^2}
$$
Write this as partial fractions, factoring the denominator as $1 - 2 z + 2 z^2 = (1 - (1 - \mathrm{i}) z) (1 - (1 + \mathrm{i}) z)$:
$$
A(z) = \frac{1 + 2 \mathrm{i}}{2} \cdot \frac{1}{1 - (1 + \mathrm{i}) z}
        + \frac{1 - 2 \mathrm{i}}{2} \cdot \frac{1}{1 - (1 - \mathrm{i}) z}
$$
This is two geometric series, so we can read off the coefficients:
$$
a_n = \frac{1 + 2 \mathrm{i}}{2} \cdot (1 + \mathrm{i})^n
        + \frac{1 - 2 \mathrm{i}}{2} \cdot (1 - \mathrm{i})^n
$$
This is the sum of a complex number and it's conjugate, so it is just twice the real part of the first one:
$$
a_n = 2 \Re\left(\frac{1 + 2 \mathrm{i}}{2} \cdot (1 + \mathrm{i})^n\right)
$$
Write the numbers in polar form:
$\begin{align}
a_n &= 2 \Re\left( \sqrt{5} \cdot \exp(\mathrm{i} \arctan 2)
                     \cdot \sqrt{2}^n \cdot \exp(\pi n \mathrm{i} / 4) \right) \\
    &= 2 \sqrt{5} \cdot \sqrt{2}^n
         \cdot \Re\left(
                     \exp\left(
                           \left(
                             \frac{n \pi}{4} + \arctan 2
                           \right) \mathrm{i}
                         \right)
                  \right) \\
    &= 2 \sqrt{5} \cdot \sqrt{2}^n
         \cdot \cos \left(
                      \frac{n \pi}{4} + \arctan 2
                    \right)
\end{align}$
