# Interesting Weighted Sum over Even Fibonacci Numbers

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?

• That sum is over even Fibonacci numbers, not odd. In any case, this is essentially a geometric sum, if you don't want to rely on the generating function.
– lulu
Commented Jul 18 at 17:02
• Commented Jul 18 at 17:31

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:

$$S_\infty=-(0+1)F_0x^0+\sum_{n=0}^\infty(n+1)F_{2n}x^n=\dfrac{\sum_{n=0}^\infty(n+1)(a^2x)^n-\sum_{n=0}^\infty(n+1)(b^2x)^n}{a-b}$$ 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 $$a+b,ab,a^2+b^2=(a+b)^2-2ab=?$$