Find $x_n$ if $x_1,x_2,\ldots,x_n$ is a set of positive numbers that satisfy $\frac{x_n+2}{2}=\sqrt{2S_n}$. 
Find $x_n$ if $x_1,x_2,\ldots,x_n$ is a set of positive numbers that satisfy $\frac{x_n+2}{2}=\sqrt{2S_n}$. 

Here $S_n$ denotes $x_1+x_2+\ldots+x_n$. Also $n\in\mathbb N_{>0}$.
It's easy to see that $x_1=2$.
I've substituted $x_n$ with $S_n-S_{n-1}$, got a quadratic equation in terms of $\sqrt{S_n}$ and found out that when $n\ge3$, we have $$\sqrt{S_n}-\sqrt{S_{n-1}}=\sqrt{2}$$
But it seems to be of no use. Any hints would be appreciated.
I could hardly find a tag this question would fit well, so you could suggest some tags as well. Thanks.

Noting that $S_n = x_n + S_{n-1}$, you have
$$
x_n^2 + 4x_n + 4 = (x_n + 2)^2 = 8S_n = 8x_n + 8S_{n-1},
$$
or
$$
x_n^2 - 4x_n + (4 - 8S_{n-1}) = 0.
$$
The solutions are
$$
x_n = \frac{1}{2}\left(4 \pm \sqrt{16 - 4(4-8S_{n-1})}\right)=2\pm\sqrt{8S_{n-1}};
$$
only the positive one is relevant, so you have
$$
x_n=2+2\sqrt{2\sum_{i=1}^{n-1}x_i}.
$$
You find that $x_1=2$, $x_2=2+2\sqrt{2\cdot 2}=6$, $x_3=2+2\sqrt{2\cdot(2+6)}=10$, $x_4=2+2\sqrt{2\cdot(2+6+10)}=14$, etc.  Pretty clearly $$x_i=4i-2.$$ The proof is by induction, using the fact that $\sum_{i=1}^{n-1}(4i-2)=2(n-1)^2$: assuming that $x_i=4i-2$ for all $i < n$,
$$
x_n=2+2\sqrt{2\cdot2(n-1)^2}=2+4(n-1)=4n-2
$$
as well.
