Find general term of recursive sequences $ x_{n+1}=\frac{1}{2-x_n}, x_1=1/2,$ Please help to solve:


*

*$ x_{n+1}=\frac{1}{2-x_n}, x_1=1/2,$ 

*$x_{n+1}= \frac{2}{3-x_n}, x_1=1/2$
I know answers, but can't figure out the solution.
The first one is obvious if you calculate first 3-5 terms by hand. But how can I get the result not by guessing, but mathematically?
Answers are:


*

*$x_n = \frac{n}{n+1}$

*$x_n = \frac{3\cdot2^{n-1}-2}{3\cdot2^{n-1}-1}$

This can be proven easily with induction. 
Without induction, for (1), define 
$$y_n := \frac{1}{1-x_n}\Rightarrow x_n = \frac{y_n-1}{y_n}$$
so $y_1=2$ and :
$$ \frac{y_{n+1}-1}{y_{n+1}}=\frac{y_n}{y_n+1}\Rightarrow y_{n+1}=y_n+1$$
From here it's clear that 
$$y_n=y_{n-1}+1=y_{n-2}+2=\ldots=y_1+n-1=n+1$$ 
and thus 
$$x_n=\frac{n+1-1}{n+1}=\frac{n}{n+1}$$
For (2) a similar idea works by defining 
$$y_n := \frac{3(x_n-1)}{x_n-2}$$
and observing that $y_n$ is a geometric progression.
