Prove that $\sum_nF_{2n+1}x^n=\frac{1-x}{1-3x+x^2}$

Question

Prove that $$\sum_nF_{2n+1}x^n=\frac{1-x}{1-3x+x^2}$$

This is a problem of generating functions. I know that $\sum_nF_{2n}x^=\frac{x}{1-3x+x^2}$ and I am guessing this is a good point to start. But I don't really know where to go from here. Could someone help me?

Answer 1

$$\sum\limits_{n=0}^\infty F_{2n}x^n=\dfrac{x}{1-3x+x^2}$$

$$=F_0+\sum_{n=0}^\infty F_{2n+2}x^{n+1}=x\sum_{n=0}^\infty F_{2n+2}x^n.$$

$$\therefore \sum_{n=0}^\infty F_{2n+2}x^n=\dfrac{\sum\limits_{n=0}^\infty F_{2n}x^n}x=\dfrac{1}{1-3x+x^2}.$$

$$\therefore\sum_{n=0}^\infty F_{2n+1}x^n=\sum_{n=0}^\infty F_{2n+2}x^n-\sum_{n=0}^\infty F_{2n}x^n=\dfrac{1-x}{1-3x+x^2},$$

since $F_{2n+1}=F_{2n+2}-F_{2n}$.