Continued fraction of $\frac{1+\sqrt{5}}{2}$

Question

I am trying to get a better understanding of continued fractions (CF) and was watching a view tutorial clips e.g. this here and looking through some stackexchange articles. Than found this article, where someone was able to get the CF of $\frac{1+\sqrt{5}}{2}$. Unfortunately he didn't post his walktrough. So I know now how I would get the CF of just $\sqrt{5}$, as you can see below, however I am not sure how to handle the whole term.

$$Int(\sqrt{5}) = 2$$ $$\sqrt{5} = 2 -2 + \sqrt{5}$$ $$= 2 + \frac{(-2 + \sqrt{5})*(2 + \sqrt{5})}{(2 + \sqrt{5})}$$ $$= 2 + \frac{(-4 - 2\sqrt{5} + 2\sqrt{5} + 5)}{2 + \sqrt{5}}$$ $$= 2 + \frac{1}{2 + \sqrt{5}}$$ $$= 2 + \frac{1}{2 + 2 + \frac{1}{2 + \sqrt{5}}} = 2 + \frac{1}{4 + \frac{1}{2 + \sqrt{5}}}$$ $$= 2 + \frac{1}{4 + \frac{1}{2 + 2 + \frac{1}{2 + \sqrt{5}}}} = 2 + \frac{1}{4 + \frac{1}{4 + \frac{1}{2 + \sqrt{5}}}}$$ $$= 2 + \frac{1}{4 + \frac{1}{4 + \frac{1}{4 + \ddots}}}$$

Can someone please help me.

Answer 1

$$ \rho = \frac{1+\sqrt{5}}2 $$ $$2 < \sqrt{5} < 3 \implies \left\lfloor \rho \right\rfloor = 1$$ $$\rho = 1 - 1 + \frac{1+\sqrt{5}}2$$ $$= 1 + \frac{\sqrt{5}-1}2$$ $$= 1 + \frac{( \sqrt{5}-1)*(1 + \sqrt{5})}{2(1 + \sqrt{5})}$$ $$= 1 + \frac{(5 + \sqrt{5} - \sqrt{5} -1 )}{2(1 + \sqrt{5})}$$ $$= 1 + \frac{4}{2(1 + \sqrt{5})}$$ $$= 1 + \frac{2}{(1 + \sqrt{5})} = 1 + \frac{1}{\rho}$$ $$= 1 + \frac{1}{1 + \frac{1}{\rho}}$$ $$= 1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{\rho}}}$$ $$= 1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + \ddots}}}$$