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)
 A: $$\frac{6+\sqrt{41}-2}{5}=\frac{10+\sqrt{41}-6}{5}=2+\frac{\sqrt{41}-6}{5}$$
with $0<\sqrt{41}-6<5$
A: Method described  by Prof. Lubin  at Continued fraction of $\sqrt{67} - 4$ 
$$  \sqrt { 41} = 6 +     \frac{  \sqrt {41} - 6 }{ 1 }  $$ 
 $$    \frac{ 1 }{   \sqrt {41} - 6 }  =  \frac{    \sqrt {41} + 6 }{5 } = 2 +  \frac{    \sqrt {41} - 4 }{5 } $$ 
 $$    \frac{ 5 }{   \sqrt {41} - 4 }  =  \frac{    \sqrt {41} + 4 }{5 } = 2 +  \frac{    \sqrt {41} - 6 }{5 } $$ 
 $$    \frac{ 5 }{   \sqrt {41} - 6 }  =  \frac{    \sqrt {41} + 6 }{1 } = 12 +  \frac{    \sqrt {41} - 6 }{1 } $$ 
Simple continued fraction tableau:
 $$ 
 \begin{array}{cccccccccccccccc}
 & & 6 & & 2 & & 2 & & 12 & & 2 & & 2 & & 12 & \\ 
 \\ 
  \frac{ 0 }{ 1 }   &   \frac{ 1 }{ 0 }   & &   \frac{ 6 }{ 1 }   & &   \frac{ 13 }{ 2 }   & &   \frac{ 32 }{ 5 }   & &   \frac{ 397 }{ 62 }   & &   \frac{ 826 }{ 129 }   & &   \frac{ 2049 }{ 320 }    \\ 
  \\ 
 & 1 & & -5 & & 5 & & -1 & & 5 & & -5 & & 1
 \end{array}
 $$ 
$$ 
 \begin{array}{cccc}
  \frac{ 1 }{ 0 }   & 1^2 - 41 \cdot 0^2 = 1 &     \mbox{digit}  &  6  \\  
  \frac{ 6 }{ 1 }   & 6^2 - 41 \cdot 1^2 = -5 &     \mbox{digit}  &  2  \\  
  \frac{ 13 }{ 2 }   & 13^2 - 41 \cdot 2^2 = 5 &     \mbox{digit}  &  2  \\  
  \frac{ 32 }{ 5 }   & 32^2 - 41 \cdot 5^2 = -1 &     \mbox{digit}  &  12  \\  
  \frac{ 397 }{ 62 }   & 397^2 - 41 \cdot 62^2 = 5 &     \mbox{digit}  &  2  \\  
  \frac{ 826 }{ 129 }   & 826^2 - 41 \cdot 129^2 = -5 &     \mbox{digit}  &  2  \\  
  \frac{ 2049 }{ 320 }   & 2049^2 - 41 \cdot 320^2 = 1 &     \mbox{digit}  &  12  \\  
 \end{array}
 $$ 
A: [This old question popped up in the feed today (years later), and I can see that there's an accepted answer. Case closed. Though, reading the question and its answers, I can't help but feeling that the problem could (should?) have been tackled differently. I'm leaving this answer in the hope it might inspire and help some future readers looking for solutions to this type of problems.]
Given that the problem statement gives the result upfront and only asks for a proof, I would have just gone the other way around. Starting from the continued fraction $X = [6;\overline {2,2,12}]$ (we remark here that it's a positive number), one can see without too much difficulty that
$$ X+6=12+\cfrac{1}{2+\cfrac{1}{2+\cfrac{1}{X+6}}} $$
And after a few mildly tedious but very easy steps, we derive
$$
\iff X+6=12+\cfrac{1}{2+\cfrac{X+6}{2X+13}} \\
\iff X+6=12+\cfrac{2X+13}{5X+32} \\
\iff X+6=\cfrac{62X+397}{5X+32} \\
\iff 5X^2+62X+192=62X+397 \\
\iff 5X^2-205=0 \\
\iff X^2-41=0 \\
$$
Since we know $X$ is positive, the result $X=\sqrt{41}$ follows.
A: 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 $6<\sqrt{41}<7$.
