<s>[INST] Continued fraction of $\frac{1+\sqrt{5}}{2}$ 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.
 [/INST] $$
\rho = \frac{1+\sqrt{5}}2
$$
$$2 < \sqrt{5} < 3 \implies \left\lfloor \rho \right\rfloor = 1$$
$$\rho = 1 - 1 + \frac{1+\sqrt{5}}2$$
$$= 1 + \frac{\sqrt{5}-1}2$$
$$= 1 + \frac{( \sqrt{5}-1)*(1 + \sqrt{5})}{2(1 + \sqrt{5})}$$
$$= 1 + \frac{(5 + \sqrt{5} - \sqrt{5} -1 )}{2(1 + \sqrt{5})}$$
$$= 1 + \frac{4}{2(1 + \sqrt{5})}$$
$$= 1 + \frac{2}{(1 + \sqrt{5})} = 1 + \frac{1}{\rho}$$
$$= 1 + \frac{1}{1 + \frac{1}{\rho}}$$
$$= 1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{\rho}}}$$
$$= 1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + \ddots}}}$$
</s>