<s>[INST] Is Binet's formula for the Fibonacci numbers exact? Is Binet's formula for the Fibonacci numbers exact?
$F_n = \frac{(1+\sqrt{5})^n-(1-\sqrt{5})^n}{2^n \sqrt{5}}$
If so, how, given the irrational numbers in it?
Thanks.
 [/INST] For linear reccurent sequence you can find expression depending on the roots of the associated polynom.
$$ F_{n+1} = F_{n} + F_{n-1} $$
is associated to 
$$ x^2 = x +1 $$
Wich has two solution, $ \frac{1 - \sqrt{5}}{2} $ and  $ \frac{1 + \sqrt{5}}{2} $ (the golden ratio, a more than interesting number)
Simple roots give a general solution of the form:
$$F_{n} = A * (\frac{1 - \sqrt{5}}{2})^{n}  + B *  (\frac{1 + \sqrt{5}}{2})^n $$
To determine A and B you have to input initial contidions: 
$$F_{0} = A + B = 0 $$
So $$  B = - A $$
and
$$F_{1} = -B * (\frac{1 - \sqrt{5}}{2}  - \frac{1 + \sqrt{5}}{2}) =1 $$
$$ B = \frac{1}{\sqrt{5}} $$
So
$$F_{n} = \frac{ (1 + \sqrt{5})^{n}  -   (1 - \sqrt{5})^n}{2^n\sqrt{5}} $$
By Solving $F_0 = i$ and $F_1 = j$ you can find the general expression of Fibonnaci sequence with starting terms i and j.
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