Continued fraction: Show $\sqrt{n^2+2n}=[n; \overline{1,2n}]$ I have to show the following identity ($n \in \mathbb{N}$):
$$\sqrt{n^2+2n}=[n; \overline{1,2n}]$$
I had a look about the procedure for $\sqrt{n}$ on Wiki, but I don't know how to transform it to $\sqrt{n^2-2n}$.
Any help is appreciated.
EDIT:
I tried the following:
$\sqrt{n^2+2n}>n$, so we get $\sqrt{n^2+2n}=n+\frac{n}{x}$, and $\sqrt{n^2+2n}-n=\frac{n}{x}$ and further $x=\frac{n}{\sqrt{n^2+2n}}$.
So we get $x=\frac{n}{\sqrt{n^2+2n}-n}=\frac{n(\sqrt{n^2+2n}+n)}{(\sqrt{n^2+2n}-n)(\sqrt{n^2+2n}+n}$ I don't know if it's right and how to go on.

HINT $\rm\  x = [\overline{1,2n}]\ \Rightarrow\  x\ = \cfrac{1}{1+\cfrac{1}{2\:n+x}}\ \iff\ x^2 + 2\:n\ x - 2\:n = 0\ \iff\ x = -n \pm \sqrt{n^2 + 2\:n} $
