I'm having some trouble with showing that a sequence $a_n$ is a solution to the recurrence relation $a_n = -3a_{n-1} + 4a_{n-2}$. (See image below). The sequence is given by $a_n = (-4)^n$.
I'm given the answer in the solutions manual, but I have absolutely no clue what is going on between step II and III. How did they get rid of the $n-1$ exponent? What did they multiply/divide/subtract/add to the equation?
\begin{align*}
-3a_{n-1}+4a_{n-2}
& =-3(-4)^{n-1}+4(-4)^{n-1} \\
& =(-4)^{n-2}\bigl((-3)(-4)+4\bigr) \\
& =(-4)^{n-2}\cdot16 \\
& =(-4)^{n-2}(-4)^2 \\
& =(-4)^n \\
& =a_n
\end{align*}
Best Answer
Since $$n-1=1+(n-2)$$ they have $$\begin{align}-3(-4)^{n-1}+4(-4)^{n-2}&=-3\cdot (-4)^{1+(n-2)}+4(-4)^{n-2}\\&=-3\cdot(-4)^1\cdot (-4)^{n-2}+4(-4)^{n-2}\\&=(-4)^{n-2}(-3\cdot (-4)+4).\end{align}$$