$F_n$ here is $n$-th fibonacci number We know that $$\sum_{n=0}^\infty \left(\sum_{k=0}^n F_kF_{n-k}\right)x^n$$ is a generating function of multiplying two G.F: $a_n =\langle F_n \rangle$ and $b_n = \langle F_n \rangle $
I was messing with this, googling, reading and stuff and i came across question about generating function of $$\sum_{n=0}^\infty \left(\sum_{k=0}^n F_{2k}F_{n-k}\right)x^n$$ (Only difference is $F_k$ and $F_{2k}$)
I know there is one nice equation $$F_{2n} = \sum_{i=0}^{n-1} F_{2i+1}$$ But when i plug it in, it becomes a little too messy and complicated.
I'd love to get some help on this question. Thanks
Okay, so to get generating function $G(x)$ of $a_n = \langle F_{2n} \rangle$ we can perform a little trick on generating function $F(x)$ of $\langle F_n \rangle$ It is, $F(x)+F(-x)$ will count 2 times even-indexed elements and 0 times odd-indexed elements, so we have to divide it by 2, so our $$G(x)=\dfrac12(F(x)+F(-x))=\frac{x^2}{(1-x-x^2)(1+x-x^2)}$$ But let's notice that $$G(x) = \sum_{n=0}^\infty F_{2n} x^{2n}$$ but we want $G(x)$ to be equal to $$\sum_{n=0}^\infty F_{2n} x^n$$ To do it, just divide every term by $x^{1/2}$ and we get $$G(x) = \sum_{n=0}^\infty F_{2n} x^n = \frac{x}{(1-\sqrt{x}-x)(1+\sqrt{x}-x)}=\frac{x}{1-3x+x^2}$$
And our answer will be $$G(x)F(x) = \sum_{n=0}^\infty (\sum_{k=0}^n F_{2k}F_{n-k}) x^n$$