Generating function for the sequence $1,1,3,3,5,5,7,7,9,9,\ldots$ The generating function for the sequence $\left\{1,1,1,1,...\right\}$ is $$1 + x + x^2 + x^3 ... = \frac{1}{1-x}$$
What is the generating function for the sequence $\left\{1,1,3,3,5,5,7,7,9,9,\dots \right\}$?

This is my attempt:
what we want to do is after the initial  $\left\{1,1,1,1,...\right\}$ function we want to add two more functions shifted two to the right such as $\left\{0,0,1,1,1,...\right\}$ and keep adding such function 2 at a time and each new two are shifted 2 more to the right then the previous one. so first would be $\left\{1,1,1,1,\dots \right\}$ then add 2 of $\left\{0,0,1,1,1,1,\dots\right\}$ then add 2 more of $\left\{0,0,0,0,1,1,1,1,\dots\right\}$ and keep doing that. this gives us the function 
$$\frac{1}{1-x}+\frac{2x^2}{1-x}+\frac{2x^4}{1-x}+...$$ or $$\frac{1}{1-x}+\frac{2x^{2k}}{1-x}$$ however $i$ do not know where to go from here, how do $i$ finish this problem?

Here is another approach.
The generating function writes $(1+x)(1+3x^2 + 5x^4 + 7x^6 + 9x^8 + \dots)$ where the second term appears as the derivative of $x+x^3 + x^5 + x^7 + x^9 + \dots$
The later being the odd part of $1+x+x^2 + x^3 + x^4 + \dots$, we have:
$$
x + x^3 + x^5 + x^7 + x^9 + \dots = \frac{1}{2}\left(\frac{1}{1-x}-\frac{1}{1+x}\right)
$$
Hence:
$$
1 + 3x^2 + 5x^4 + 7x^6 + 9x^8 + \dots = \frac{1}{2}\left(\frac{1}{(1-x)^2}+\frac{1}{(1+x)^2}\right)
$$
Finnaly:
$$
\frac{1+x}{2}\left(\frac{1}{(1-x)^2}+\frac{1}{(1+x)^2}\right) = \frac{x^2+1}{(x-1)^2(1+x)}
$$
