Show recursion in closed form I've got following sequence formula:
$ a_{n}=2a_{n-1}-a_{n-2}+2^{n}+4$
where $ a_{0}=a_{1}=0$
I know what to do when I deal with sequence in form like this:
$ a_{n}=2a_{n-1}-a_{n-2}$
  -  when there's no other terms but previous terms of the sequence.
Can You tell me how to deal with this type of problems? 
What's the general algorithm behind solving those?

One approach that works to get to the final form is to take the formal power series
$$f(x) = \sum_{n=0}^{\infty} a_n x^n$$
and try and rewrite it in terms of itself. Applying the initial conditions where necessary:
\begin{align}
f(x) &= \sum_{n=0}^{\infty} a_n x^n\\
&= a_0 + a_1 x +\sum_{n=2}^{\infty} \left(2a_{n-1}-a_{n-2} + 2^n + 4\right) x^n\\
%
&= 2\sum_{n=2}^{\infty}a_{n-1}x^n - \sum_{n=2}^{\infty}a_{n-2}x^n + \sum_{n=2}^{\infty}2^nx^n + 4\sum_{n=2}^{\infty}x^n\\
%
&= 2\sum_{n=1}^{\infty}a_nx^{n+1} - \sum_{n=0}^{\infty}a_n x^{n+2} + \sum_{n=0}^{\infty}(2x)^{n+2} + 4\sum_{n=0}^{\infty}x^{n+2}\\
%
&= 2x\left(\sum_{n=0}^{\infty}a_nx^n - a_0\right) - x^2\sum_{n=0}^{\infty}a_n x^n + 4x^2\sum_{n=0}^{\infty}(2x)^n + 4x^2\sum_{n=0}^{\infty}x^n\\
%
&= 2xf(x) - x^2f(x) + 4x^2 \frac{1}{1-2x} + 4x^2 \frac{1}{1-x}\\
%
&= (2x-x^2)f(x) + 4x^2\frac{(1-x)+(1-2x)}{(1-2x)(1-x)}\\
%
&= (2x-x^2)f(x) + \frac{4x^2(2-3x)}{(1-2x)(1-x)}
\end{align}
Solving for $f(x)$:
$$f(x) = \frac{4x^2(2-3x)}{(1-2x)(1-x)(1-2x+x^2)} = \frac{4x^2(2-3x)}{(1-2x)(1-x)^3}$$
Applying partial fraction expansion and using the well-known result that 
$$\sum_{n=0}^{\infty} (n+1)(n+2)\dotsb(n+m)x^n
= \frac{d^m}{dx^m}\left(\frac{1}{1-x}\right)
= \frac{m!}{(1-x)^{m+1}}$$
we get
\begin{align}
\sum_{n=0}^{\infty} a_n x^n &=
\frac{4x^2(2-3x)}{(1-2x)(1-x)^3}\\
&= \frac{4}{1-2x} + \frac{4}{1-x} - \frac{12}{(1-x)^2} + \frac{4}{(1-x)^3}\\
&= 4\frac{1}{1-2x} + 4\frac{1}{1-x} - 12\frac{1!}{(1-x)^{1+1}} + 2 \frac{2!}{(1-x)^{2+1}}\\
&= 4 \sum_{n=0}^{\infty} (2x)^n + 4 \sum_{n=0}^{\infty} x^n - 12 \sum_{n=0}^{\infty} (n+1) x^n + 2 \sum_{n=0}^{\infty} (n+1)(n+2) x^n\\
&= \sum_{n=0}^{\infty} \left(4\cdot 2^n + 4 - 12(n+1) + 2(n+1)(n+2)\right)x^n\\
&= \sum_{n=0}^{\infty} \left(2^{n+2} + 2n^2 - 6n -4\right)x^n
\end{align}
Equating coefficients, we find
$$a_n = 2^{n+2} + 2n^2 - 6n -4$$
