Is there some strategy to find the general term of a recursive sequence? Given $(a_{n})_{n \in \mathbb{N}}$,
$a_{0} = 2, a_{1} = 4, a_{n+2} = 4a_{n+1} - 3a_{n}$, 
is there any way to find the general term? I reckoned that every $a_{i}$ is a factor of 2, and then the series becomes 1, 2, 5, 14, 41, ..., and obviously that's $a_{i} = 3(a_{i-1}) - 1$ (if $i≠0$) but even then I just can't get rid of these damn recursive clauses.
Any help? Is there some way to tackle this kind of problems?

Writing each pair of consecutive terms as a linear combination of the previous pair of consecutive terms,
$$\pmatrix{a_{n}\\a_{n-1}} = \pmatrix{4&-3\\1&0}\pmatrix{a_{n-1}\\a_{n-2}}= \cdots = \pmatrix{4&-3\\1&0}^{n-1}\pmatrix{a_{1}\\a_{0}}$$
The question becomes how to represent the matrix power in close form, which is by diagonalisation.
$$\begin{align*}
\pmatrix{4-\lambda&-3\\1&0-\lambda} &= 0\\
(4-\lambda)(-\lambda) +3 &= 0\\
(\lambda-3)(\lambda-1) &= 0
\end{align*}$$
(This characteristic equation has the same roots as the ones given in the other answers.)
For $\lambda = 3$, $$\pmatrix{4&-3\\1&0}\pmatrix{3\\1} = 3\pmatrix{3\\1}$$
For $\lambda = 1$, $$\pmatrix{4&-3\\1&0}\pmatrix{1\\1} = \pmatrix{1\\1}$$
So
$$\begin{align*}
\pmatrix{4&-3\\1&0}^{n-1} &= \pmatrix{3&1\\1&1}\pmatrix{3^{n-1}&0\\0&1}\pmatrix{3&1\\1&1}^{-1}\\
&= \frac12\pmatrix{3&1\\1&1}\pmatrix{3^{n-1}&0\\0&1}\pmatrix{1&-1\\-1&3}\\
a_n &= \pmatrix{1&0}\pmatrix{a_{n}\\a_{n-1}}\\
&= \pmatrix{1&0}\pmatrix{4&-3\\1&0}^{n-1}\pmatrix{a_{1}\\a_{0}}\\
&= \pmatrix{1&0}\pmatrix{3&1\\1&1}\pmatrix{3^{n-1}&0\\0&1}\pmatrix{3&1\\1&1}^{-1}\pmatrix{4\\2}\\
&= \pmatrix{1&0}\pmatrix{3&1\\1&1}\pmatrix{3^{n-1}&0\\0&1}\pmatrix{1&-1\\-1&3}\pmatrix{2\\1}\\
&= \pmatrix{3&1}\pmatrix{3^{n-1}&0\\0&1}\pmatrix{1\\1}\\
&= 3^n+1
\end{align*}$$
