Doing some reading when I come across this: " ...clearly $$\sum_{n=1}^{\infty}\frac{(n+1)F_{2n}}{3^{n+1}} = 9$$ where $F_n$ is the nth Fibonacci number evaluates to $9$. We derive this solution from the general case $\sum_{n=1}^{\infty}(n+1)F_{2n}x^{n+1}$ where the general solution is left as an exercise."
Does anyone know how to approach sums of this form?
Multiple approaches come to mind. The generating function for the Fibonnaci numbers under the definition
$$ F_0=0,\;\;\;\;F_1=1,\;\;\;\;F_{n+2}=F_{n+1}+F_n $$
is
$$ \sum_{n=0}^\infty F_nx^n=\frac{x}{1-x-x^2} $$
We can cancel all of the odd terms by performing
$$ \sum_{n=0}^\infty F_{2n}x^{2n}=\frac{1}{2}\left(\sum_{n=0}^\infty F_nx^n+\sum_{n=0}^\infty F_n(-x)^n\right) \\ =\frac{1}{2}\left(\frac{x}{1-x-x^2}-\frac{x}{1+x-x^2}\right)=\frac{x^2}{1-3x^2+x^4} $$
Replacing $x^2\to x$,
$$ \sum_{n=0}^\infty F_{2n}x^n=\frac{x}{1-3x+x^2} $$
We multiply by $x$ and differentiate to get
$$ \sum_{n=0}^\infty(n+1)F_{2n}x^n=\frac{d}{dx}\left[\frac{x^2}{1-3x+x^2}\right]=\frac{2x-3x^2}{(1-3x+x^2)^2} $$
Noting $F_0=0$, we drop the first term and multiply by $x$ to get
$$ \sum_{n=1}^\infty(n+1)F_{2n}x^{n+1}=\frac{2x^2-3x^3}{(1-3x+x^2)^2} $$
This is your general form, and note that $x=1/3$ recovers that the sum is $9$.
Hint:
Like Prove that for every $ n \in \mathbb{N} \cup \{0\}$, $F_{n+5} + 2F_n$ is divisible by $5$.
and using How can I evaluate $\sum_{n=0}^\infty(n+1)x^n$?,
$$S_\infty=-(0+1)F_0x^0+\sum_{n=0}^\infty(n+1)F_{2n}x^n=\dfrac{\sum_{n=0}^\infty(n+1)(a^2x)^n-\sum_{n=0}^\infty(n+1)(b^2x)^n}{a-b}$$ where $a,b(a>b)$ are the roots of $t^2-t-1=0\implies a+b=1,ab=-1$
If $|a^2x|,|b^2x|<1,$
$$S_\infty=\dfrac{\dfrac1{(1-a^2x)^2}-\dfrac1{(1-b^2x)^2}}{a-b} =\dfrac{(1-b^2x)^2-(1-a^2x)^2}{(a-b)(1-(a^2+b^2)x+x^2(ab)^2)^2}$$
Now $(1-b^2x)^2-(1-a^2x)^2=x(a-b)(a+b)(2-(a^2+b^2)x)$
Finally cancel out $a-b(\ne0)$ and replace the values of $a+b,ab,a^2+b^2=(a+b)^2-2ab=?$
