Discrete math: $a_n=14a_{n−1}−33a_{n−2}$ for $n≥3$ with initial conditions $a_0=−24,a_1=−200$ Can someone please help me understand this? This is what I got but it isn't correct.
$a_n=14a_{n−1}−33a_{n−2}$ for $n≥3$ with initial conditions $a_0=−24,a_1=−200$.
Solve for $a_n$.
$t^2 = -14t+33$
$r_1=-11$
$r_2= -3$
$n_0 = M(-11)^0 + N(-3)^0 = -24$
$n_1 = M(-11)^1 + N(-3)^1 = -200$
By algebra, $m = 34 , n = -58$, so
$a_n= 34(-11)^n - 58(-3)^n$.
 A: Just a sign error in the setup.  You suppose that $a_n = t^n$ is the general rule, and plugging into the recurrence relation (for $n \geq 3$) gives 
$$t^n = 14 t^{n-1} - 33 t^{n-2},$$
or, putting everything on one side,
\begin{align}
t^n - 14 t^{n-1} + 33 t^{n-2} &= 0\\
t^{n-2}(t^2 - 14 t + 33) & = 0\\
\end{align}
[I trust the nature of the sign error is already clear.]
Thus, for $n \geq 3$, we have that 
\begin{align}
t^{n-2} & = 0 && \text{or} && t^2 - 14t + 33 && = 0\\
t & = 0 && \text{or} && (t-11)(t-3) && = 0\\
t & = 0 && \text{or} && t -11 && = 0 && \text{or} && t-3 && = 0\\
t & = 0 && \text{or} && t && = 11 && \text{or} && t && = 3\\
\end{align}
and we see that the roots are $0$, $\mathbf{+}11$, and $\mathbf{+}3$.  I presume that $t = 0$ ends up having no role (since $0^n$ is just $0$ and it is just extra hassle to add it when it does not help).
A: Here is a differential equations solution:
$f(x)=14f(x-1)-33f(x-2)$
With a bit of work we can find,
$f(x)=c_{1}(7-\sqrt82)^x+c_{2}(7+\sqrt82)^x$.
We just need to find the constants $c_{1}$ and $c_{2}$. Luckily, you've provided us with initial conditions I can use to find $c_1$ and $c_2$.
$a_0=-24$
$a_1-200$
Or,
$f(0)=-24$
$f(1)=-200$
To solve this equation we must solve, 
$c_{1}+c_{2}=-24$
And,
$-200=(7-\sqrt82)c_{1}+(7+\sqrt82)c_{2}$
These equations as tedious to solve so I plugged them into my scientific calculator and apparently:
$c_{1}=-2(3\sqrt82-4)\sqrt\frac{2}{\sqrt41}$
And,
$c_{2}=-12-8\sqrt\frac{2}{\sqrt41}$
If you plug the constants into the sequence you'll get your answer. By the way, $f(x)=a_{n}$.
