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.
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

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