continued fraction of $\sqrt{41}$ 
Show that $\sqrt{41} = [6;\overline {2,2,12}]$

here's my try:
$$\sqrt{36}<\sqrt{41}<\sqrt{49}\implies6<\sqrt{41}<7\implies\lfloor\sqrt{41}\rfloor=6$$
$$\sqrt{41}=6+\sqrt{41}-6=6+\frac{1}{\frac{1}{\sqrt{41}-6}}$$
$$\frac{1}{\sqrt{41}-6}=\frac{\sqrt{41}+6}{41-36}=\frac{\sqrt{41}+6}{5}=\frac{12+\sqrt{41}-6}{5}=2+\frac{\sqrt{41}-4}{5}$$
So far,
$$\sqrt{41}=6+\frac{1}{2+\frac{\sqrt{41}-4}{5}}=6+\frac{1}{2+\frac{1}{\frac{5}{\sqrt{41}-4}}}$$
But, $$\frac{5}{\sqrt{41}-4}=\frac{5(\sqrt{41}+4)}{41-16}=\frac{\sqrt{41}+4}{5}=\frac{6+\sqrt{41}-2}{5}=\color{red}{1}+\frac{\sqrt{41}-1}{5}$$
It suppose to be $2$ and not $1$. 
Where is the mistake? (I triple-checked and it seems fine to me)

Given $(\sqrt{41}+4)/5$, render $6<\sqrt{41}<7$ which you used at the start.  Then your fraction lies between $(6+4)/5$ and $(7+4)/5$ showing the integral part is $2$.
You should be able to get the integer part at every stage just from $6<\sqrt{41}<7$.
