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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.

My answer:

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.

Thank you!

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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
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I don't know LaTeX, which is why I included both links to show the problem and answer –  stevenmadden Apr 14 '13 at 19:52
    
@jtm22, go to the FAQ section for directions on using LaTeX in this site. As it is your question is very hard to understand. –  DonAntonio Apr 14 '13 at 21:10
    
Please make sure that I copied the images correctly. –  Brian M. Scott Apr 15 '13 at 7:57
    
The defacing of this user's questions continues... –  Did May 6 '13 at 19:33

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