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I was doing some question on Fibonacci Sequence and came across this formula $$\large F_n = \frac{\left(\frac{1+\sqrt5}{2}\right)^n - \left(\frac{1-\sqrt5}{2}\right)^n}{\sqrt5}$$ I started thinking “Is there a better way to write this formula? Because this seems ugly, look at the denominator, it has surd and the numerator is also not good for easy calculation”
So I did some calculations and reformed it in another way $$ \large F_n^2 = \left\{ \begin{array}{ll} \dfrac{\left\lfloor \left(\frac{3+\sqrt5}{2}\right)^n \right\rfloor-1} {5}; \;\text{n is even}\\\\ \dfrac{\left\lfloor \left(\frac{3+\sqrt5}{2}\right)^n \right\rfloor+3}{5} ; \;\text{n is odd} \end{array} \right. $$ How can we further simplify it? I want to cancel $5$ in the denominator and need to prove that these two expressions are perfect squares

user1150809
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  • I think for Fibonacci numbers greater than 10, you only need the first term in the numerator, keep the same denominator, then take the floor. – TurlocTheRed Mar 22 '24 at 05:27
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    I don't know that that's actually a simplification. – Arthur Mar 22 '24 at 06:22
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    Note that $\left|\dfrac{1-\sqrt{5}}{2}\right|<1$ and $\sqrt{5}>2$, so $\left|\dfrac{1}{\sqrt{5}}\left(\dfrac{1-\sqrt{5}}{2}\right)^n\right|<\dfrac{1}{2}$ for $n\ge 0$, and therefore $F_n$ is the closest integer to $\dfrac{1}{\sqrt{5}}\left(\dfrac{1+\sqrt{5}}{2}\right)^n$. – Alexander Burstein Mar 22 '24 at 07:03

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