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.

Because $n^2+2n=(n+1)^2-1$, the leading term in the continued fraction expansion will be $n$.  Multiplying by a form of $1$, we have
$$\sqrt{n^2+2n}-n=\frac{2n}{n+\sqrt{n^2+2n}}$$
 and $1<\frac{n+\sqrt{n^2+2n}}{2n}<2,$ so the next term will be $1$.  Since $\frac{n+\sqrt{n^2+2n}}{2n}-1=\frac{\sqrt{n^2+2n}-n}{2n}$ and we can again rationalize the denominator to simplify $\frac{2n}{\sqrt{n^2+2n}-n}=n+\sqrt{n^2+2n}$.  Because $n<\sqrt{n^2+2n}<n+1$, the next term in the continued fraction expansion is $2n$, and subtracting 2n yields $\sqrt{n^2+2n}-n$, at which point the pattern continues.
In the end, it all boils down to the useful trick that $\frac{1}{\sqrt{a}-\sqrt{b}}=\frac{\sqrt{a}+\sqrt{b}}{a-b}$, which can be viewed as a application of $a^2-b^2=(a+b)(a-b)$.  Judging from your edit, you are aware of this idea, but you just didn't carry out the necessary algebraic simplifications to see that it actually leads somewhere.
