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.

The first few convergents of $[3;\overline{1,6}]$ are $3,4,\frac{27}7,\frac{31}8,\frac{213}{55}$, and $\frac{244}{63}$; $\frac{213}{55}=3.8\overline{72}$, and $\frac{244}{63}$ is a little over $3.873$, so $[3;\overline{1,6}]$ is clearly not $\sqrt{3^2-2\cdot3}=\sqrt3$.
In fact, $n^2-2n=(n-1)^2-1$, so the integer part and first convergent of $\sqrt{n^2-2n}$ will be $n-2$. But $\sqrt{n^2-2n}$ isn’t $[n-2;\overline{1,2n}]$, either, since, as you can see here, $\sqrt3=[1;\overline{1,2}]$.
$\sqrt8=[2;\overline{1,4}]$, so your identity should probably be $\sqrt{n^2-2n}=[n-2;\overline{1,2(n-2)}]$ for $n\ge 3$.
Now let $m=n-2$ and consider the continued fraction $x=[m;\overline{1,2m}]$.
$$\begin{align*}
x+m&=[2m;\overline{1,2m}]\\
&=2m+\frac1{1+\frac1{[2m;\overline{1,2m}]}}\\
&=2m+\frac1{1+\frac1{x+m}}\\
&=2m+\frac{x+m}{x+m+1}\\
&=\frac{(2m+1)x+2m^2+3m}{x+m+1}\;,
\end{align*}$$
and you can finish it off by solving for $x$ in terms of $m$ (and then in terms of $n$).
If the identity was supposed to have $\sqrt{n^2+2n}$ on the lefthand side, the same basic approach will work, though the details will obviously be different, and you won’t need $m$.
Added: I see that you have edited the problem statement to make the lefthand side $\sqrt{n^2+2n}$; I’ll leave my solution to the original version as an extended hint for the corrected version.
