Discrete math help : $a_n=−3a_{n−1}$ for $n\geq 2$ with the initial condition $a_1=−12$. I am thoroughly confused , and am not sure how to solve this.
Could someone please walk me through this.
I have an idea how to do it with an $a_0$ and an $a_1$ , but not this one.
 A: Hint: You can always start by writing out the first few terms...
$a_{1}=-12$
$a_{2}=-3a_{1}=-3(-12)=36$
$a_{3}=-3a_{2}=-3(36)=-108$
and so on.  Sometimes it is helpful to not actually do all the multiplication.  For example,
$a_{1}=-12$
$a_{2}=(-3)(-12)$
$a_{3}=(-3)a_{2}=(-3)(-3)(-12)=(-3)^{2}(-12)$
$a_{4}=(-3)a_{3}=(-3)(-3)(-3)(-12)=(-3)^{3}(-12)$
$a_{5}=(-3)a_{4}=(-3)(-3)^{3}(-12)=(-3)^{4}(-12)$
and then, look for a pattern... Pay particular attention to the subscript on the $a$ and the exponent of $(-3)$; you can even make things a little simpler by factoring a $(-3)$ out of the $(-12)$.
A: I assume you want a formula for $a_n$.
If $a_1=-12$ we can say that $a_0=4$, since that preserves the relationship. I find that finding a formula is easier if I start from $n=0$ rather than $n=1$: it is usually easier and ends up with an easier formula, as in this case.
Each time we increase $n$ we multiply $a_n$ by $-3$. That gives us the idea that $a_n$ is related to powers of $-3$, and we end up with
$$a_n=4(-3)^n$$
That is true for all natural numbers, $n\ge 0$.
A: I always use differential equations for these problems.
We really have,
$f(x)=-3f(x-1)$.
If we do a bit of differential equations magic we'll get:
$f(x)=c_{1}(-3)^{x-1}$.
This is equivalent  to $a_n=c_{1}(-3)^{n-1}$.
We know have to solve for $c_1$ which is easy since this is a differential equation of first order and you've already provided a initial condition.
We have,$c_{1}(-1)^{0}=-12$ therefore $c_{1}=-12$.
Therefore, $a_n=-12(-3)^{n-1}$.
EDIT: Which is equivalent to $a_{n}=4(-3)^n$
