Find the closed form for the following: $$T(n) = \begin{cases} 1\quad &\text{ if } n = 0 \\ 4\quad &\text{ if } n = 1 \\ 4T(n-1) - 4T(n-2) & \text{ if } n > 1 \end{cases}$$
Usually I would create a general formula based on $k$ repetitions, then solve for $k$ based on a value of $n$ and substitute. From there it is usually easy to get a formula for the equation in terms of $n$. But I don't see how this would work here.
We have $$T_n=4T_{n-1}+4T_{n-2}$$
Doing substitution $T_n=\alpha^n$ we get
$$\alpha^2-4\alpha+4=0 \implies (\alpha-2)^2=0$$
Thus, the general form of $T_n$ is given by
$$T_n=A\cdot 2^n+B\cdot n2^n$$
We know that $T_0=1$ and $T_1=4$ so finally
$$T_n=2^n+n2^n$$