Question: Find the generating functions of the sequences
2, 1, 2, 1, 2, 1, . . . 

I get $\frac{1}{1+x} + \frac{1}{1-x} = \frac{2}{1-x^2}$

But the solution ends up with $\frac{2}{1-x^2} + \frac{x}{1-x^2} = \frac{2+x}{1-x^2}$.
The solutions starts with $\sum\_{n\ge 0} (2)x^{2n}+\sum\_{n\ge 0} (1) x^{2n+1}$

I couldn't come up with anything like that. I feel like I'm confused with something.

Answer: 
As $\frac{1}{1-x}$ is the generating series for $1,1,1,\dots$ and $\frac{1}{1-x}$ is the generating series for $1,-1,1,-1,\dots$, the series for which you are calculating the generating function for is $2,0,2,0,\dots$.

But we can get the result you want by noticing that $2,1,2,1,\dots$ is actually $\frac{3}{2},\frac{3}{2},\frac{3}{2},\dots$ plus $\frac{1}{2},-\frac{1}{2},\frac{1}{2},-\frac{1}{2},\dots$.

We know the first series has generating function $\frac{3/2}{1-x}$ and the second has generating function $\frac{1/2}{1+x}$. Thus, the function you are looking for is$$\frac{\frac{3}{2}}{1-x}+\frac{\frac{1}{2}}{1+x}=\frac{\frac{3}{2}(1+x)+\frac{1}{2}(1-x)}{(1-x)(1+x)}=\frac{2+x}{1-x^2}$$
which is the result you were given.
And if you want to generate
$(a\_i)\_{i=1}^m$ repeatedly,
the i-th term is generated by
$a\_ix^i$,
and for this to repeat every $m$
this needs
$\dfrac{a\_ix^i}{1-x^m}
$
so the final result is
$\sum\_{i=1}^m\dfrac{a\_ix^i}{1-x^m}
$.

In your case,
with $(1, 2)$
repeated,
this is
$\dfrac{x}{1-x^2}+\dfrac{2x^2}{1-x^2}
=\dfrac{x+2x^2}{1-x^2}
$.

This starts the first term at $x^1$.
To start it at the constant term,
just reduce the exponents
of the $x^i$ by one giving
$\sum\_{i=1}^m\dfrac{a\_ix^{i-1}}{1-x^m}
$,
so this gives,
in this case
$\dfrac{1}{1-x^2}+\dfrac{2x}{1-x^2}
=\dfrac{1+2x}{1-x^2}
$.