Showing that a sequence $a_n=(-4)^n$ is a solution of the recurrence relation $a_n = -3a_{n-1} + 4a_{n-2}$

Question

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*}

Answer 1

Since $$n-1=1+(n-2)$$ they have $$\begin{align}-3(-4)^{n-1}+4(-4)^{n-2}&=-3\cdot (-4)^{1+(n-2)}+4(-4)^{n-2}\\&=-3\cdot(-4)^1\cdot (-4)^{n-2}+4(-4)^{n-2}\\&=(-4)^{n-2}(-3\cdot (-4)+4).\end{align}$$

Answer 2

He didn't get "rid of it". Remember that $$(-4)(-4)^{-1} = 1$$

Then $$ -3(-4)^{n-1} + 4(-4)^{n-2} = -3(-4)^{n-1}(-4)(-4)^{-1} + 4(-4)^{n-2} = $$

$$ -3(-4)^{n-2}(-4) + 4(-4)^{n-2}$$

And step III follows by the distributive of multiplication.

Answer 3

So we start with $a_n=(-4)^n$ for all $n$, which means that $a_{n-1}=(-4)^{n-1}$ and $a_{n-2}=(-4)^{n-2}$.

We then substitute these two values into the recurrence $a_n=-3a_{n-1}+4a_{n-2}$ to obtain $$a_n=-3\times(-4)^{n-1}+4\times(-4)^{n-2} $$

The rest is just dealing with $a^r\cdot a^s=a^{r+s}$, where $a=-4$ and $bc+bd=b(c+d)$ where $b$ is a power of $-4$. You just have to be careful about the signs.