recurrence relation where $c_n = c_{n-1} + 2c_{n-2}$ 
A sequence $(c_n)$ is defined recursively as follows: $c_0 = 1, c_1 = 1, $ and $c_n = c_{n-1} + 2c_{n-2}$ for $n\geq 2$. We use $[x^n]g(x)$ to denote the coefficient of $x^n$ of the polynomial $g(x).$ Show that $c_{2n} = [x^{2n}] \dfrac{1}2\left(\dfrac{1}{1-x-2x^2}+\dfrac{1}{1+x-2x^2}\right)$ and that $\sum_{n\geq 0} c_{2n}x^n = \dfrac{1-2x}{1-5x+4x^2}.$ From this, one can deduce that $c_{2n} = 5c_{2n-2} - 4c_{2n-4}.$ Obtain a similar equation for $c_{2n}, c_{2n-4}$ and $c_{2n-8}.$

I know that $\dfrac{1}2\left(\dfrac{1}{1-x-2x^2}+\dfrac{1}{1+x-2x^2}\right) = \dfrac{1-2x^2}{1-5x^2+4x^4}$ and so if I show that $c_{2n} = [x^{2n}] \dfrac{1-2x^2}{1-5x^2+4x^4},$ I can replace $x^2$ with $x$ and get that $\sum_{n\geq 0} c_{2n}x^n = \dfrac{1-2x}{1-5x+4x^2}$. But I'm not sure how to show that $c_{2n} = [x^{2n}] \dfrac{1-2x^2}{1-5x^2+4x^4}.$ I don't think I'll need to compute the exact coefficient and it does not seem useful to manipulate the recurrence equation by substituting $n$ with $2n$. I tried showing that $(1-5x^2+4x^4) \sum_{n\geq 0} c_{2n} x^{2n} = 1-2x^2.$ Matching coefficients results in $c_0+(c_2-5c_0)x^2 + \sum_{n\geq 0} (c_{2n}-5c_{2n-2}-2c_{2n-4})x^{2n} = 1-2x^2,$ but it seems I'd need to prove something like $c_{2n}-5c_{2n-2}-2c_{2n-4}.$ I think figuring out how to come up with $\dfrac{1-2x}{1-5x+4x^2}$ should help me obtain a similar equation relating $c_{2n}, c_{2n-4}, c_{2n-8}$

Using generating functions, I obtained the expression for $G(z) = \sum_n c_n z^n$:
$$
G(z) = \frac{c_0(1-z)+c_1z}{(1+z)(1-2z)}
$$
Now you need to expand the expression with partial fractions, you will obtain on RHS two expressions of the form
$$
G(z) = \lambda_1\sum_{z=0}^{\infty}(-1)^kz^k + \lambda_1\sum_{z=0}^{\infty}(-s_1)^kz^k
$$
By taking the coefficients for the term $z^n$ you will get your closed-form expression. Here you will need to find the constants $\lambda_1, \lambda_2$ using partial fractions, and $s_1$ by using Generalized binomial coefficient for $\frac{1}{1-2z}$. Can you handle from here?
EDIT: In the partial fraction step it's better to group $c_0 + (c_1-c_0)z$
