# Is this theorem proof correct?

I'm trying to prove this theorem:

Let $c_1$ and $c_2$ be real numbers with $c_2 \ne 0$. Suppose that $r^2 − c_1r − c_2 = 0$ has only one root $r_0$. A sequence $\{a_n\}$ is a solution of the recurrence relation $a_n = c_1a_{n−1} +c_2a_{n−2}$ if and only if $a_n = \alpha_1r_0^n+ \alpha_2nr_0^n$, for $n = 0, 1, 2, \dots$ , where $\alpha_1$ and $\alpha_2$ are constants.

Proof. Suppose $a_n=\alpha_1r_0^n+ \alpha_2nr_0^n$. By definition, $r_0$ is the only root of $r^2-c_1r-c_2$, so $r_0^2=c_1r_0+c_2$. Moreover, since it’s the only root, it’s also a root of the derivative, so $2r_0=c_1$. Then \begin{align*}c_1a_{n-1}+c_2a_{n-2}&=c_1[\alpha_1r_0^{n-1}+\alpha_2(n-1)r_0^{n-1}]+c_2[\alpha_1r_0^{n-2}+\alpha_2(n-2)r_0^{n-2}]\\&=\alpha_1r_0^{n-2}[c_1r_0+c_2]+\alpha_2r_0^{n-2}[c_1(n-1)r_0+c_2(n-2)]\\&=\alpha_1r_0^n+\alpha_2r_0^{n-2}[n(c_1r_0+c_2)-c_1r_0-2c_2]\\&=\alpha_1r_0^n+\alpha_2nr_0^n+\alpha_2r_0^{n-1}[c_1-2r_0]\\&=a_n\;,\end{align*} so $a_n$ satisfies the recurrence relation.
If $a_0=C_0$ and $a_1=C_1$, then $\alpha_1=C_0$, and $C_1=(\alpha_0+\alpha_1)r_0$, so $\alpha_1=\frac{C_1-C_0r_0}{r_0}$. By Math $55$, this is a unique solution, and so we’re done.
This is very hard to understand. Please format it in standard LaTeX so we can know what you really mean. For example, why is c2 = 0? Also, by "r2" do you mean $r^2$? – marty cohen Apr 14 '13 at 19:51