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?

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$\ds{a_{n} = 2a_{n - 1} - a_{n - 2} + 2^{n} + 4\,,\qquad a_{0} = a_{1} = 0:\ {\large ?}}$.

Note that
$\ds{a_{n} - a_{n - 1} = a_{n - 1} - a_{n - 2} + \pars{2^{n} + 4}}$. Then,
\begin{align}
\underbrace{\sum_{k = 2}^{n}\pars{a_{k} - a_{k - 1}}}
_{\ds{a_{n} - a_{1} = a_{n}}}\ &\ =\
\underbrace{\sum_{k = 2}^{n}\pars{a_{k - 1} - a_{k - 2}}}
_{\ds{a_{n - 1} - a_{0} = a_{n - 1}}}\ +\
\underbrace{\sum_{k = 2}^{n}\pars{2^{k} + 4}}_{\ds{2^{n + 1} + 4n - 8}}
\\[5mm] \implies a_{n} - a_{n - 1} & = 2^{n + 1} + 4n - 8
\end{align}

Similarly,
\begin{align}
\overbrace{\sum_{k = 2}^{n}\pars{a_{k} - a_{k - 1}}}^{\ds{a_{n} - a_{1} = a_{n}}}\ & \ =\
\sum_{k = 2}^{n}\pars{2^{k + 1} + 4k - 8} =
\sum_{k = 1}^{n - 1}\pars{2^{k + 2} + 4k - 4}
\\[5mm] & =
2^{3}\,{2^{n - 1} - 1 \over 2 - 1} + 4\,{\pars{n - 1}n \over 2} - 4\pars{n - 1}
\end{align}

$$
\bbx{a_{n} = 2^{n + 2} + 2n^{2} - 6n - 4\,,\qquad n = 0,1,2,\ldots}
$$
