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 an alternative to the other answers, you can simply use a variant of one function, that being

$$\frac{1}{1-x^2}=\sum\_{n=0}^{\infty} x^{2n}$$

If you consider that the $2$'s are even indexed (assuming in the sequence $\{a\_n\}$ the first term is $a\_0$) and the $1$'s are odd indexed, you have that 

$$\frac{2}{1-x^2}$$

will generate the $2$'s and 

$$\frac{x}{1-x^2}$$

will generate the $1$'s. This is because 

$$\frac{1}{1-x^2}=1+x^2+x^4+...$$

and so

$$\frac{x}{1-x^2}=x(1+x^2+x^4+...)=x+x^3+x^5+...$$

Thus, you sequence will be generated by

$$f(x)=\frac{2+x}{1-x^2}$$
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}
$.