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The following is from p.4 of https://www.math.ucdavis.edu/~hunter/intro_analysis_pdf/ch3.pdf

The terms in the Fibonacci sequence are uniquely determined by the linear difference equation

$$F_n − F_{n−1} − F_{n−2} = 0, n ≥ 3$$

with the initial conditions $F_1 = 1, F_2 = 1$.

We see that $F_n = r^n$ is a solution of the difference equation if $r$ satisfies $r^2 − r − 1 = 0$ which gives $r = \phi$ or −$\dfrac{1}{\phi}$ where $\phi = \dfrac{1 + \sqrt{5}}{2}\approx 1.61803$.

I'm unable to see why $F_n = r^n$ is a solution of the difference equation if $r$ satisfies $r^2-r-1=0$.

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    $\begingroup$ Do you mean $F_n$ not $F^n$? If so, plugging in $F_n=r^n$, we get $r^n-r^{n-1}-r^{n-2}=r^{n-2}(r^2-r-1)=0$ and clearly $r\ne0$. $\endgroup$ – TheSimpliFire Dec 28 '18 at 19:12
  • $\begingroup$ See the answer at this duplicate. $\endgroup$ – Dietrich Burde Dec 28 '18 at 19:15
  • $\begingroup$ @TheSimpliFire: Yes, I meant $F_n$. $\endgroup$ – K.M Dec 28 '18 at 19:28
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Hint: Multiply $r^2 − r − 1 = 0$ by $r^{n-2}$.

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  • $\begingroup$ Multiplying $r^2 -r-1=0$ by $r^{n-2}$, I get that $r^{n}-r^{n-1}-r^{n-2}=0$, but I'm not sure how I would get that $F_{n}=r^{n}$ without first assuming that $F_n=r^{n}$ and then plugging this into $r^{n}-r^{n-1}-r^{n-2}=0$ $\endgroup$ – K.M Dec 29 '18 at 21:14
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Usually, the complete solution to a difference equation with two distinct roots in the characteristic equation is in the form $$A{r_1}^n + B{r_2}^n$$.

In your case, the distinct roots for $r_1$ and $r_2$ are $\phi = \dfrac {1+\sqrt{5}}{2}$ and $-\dfrac {1}{\phi} = \dfrac {1-\sqrt{5}}{2}$. Thus the complete solution is $$F_n = \dfrac {1}{\sqrt {5}} \left (\phi^n - \dfrac {1}{\phi}^n \right)$$

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