Is there any way to calculate
$\sum_{n=1}^{\infty} \frac{F_{2n}}{2^{2n}}$
where $F_n$ is the nth Fibonacci number, $F_1=0$, $F_2=1$, $F_n = F_{n-1} + F_{n-2}$
Nth Fibonacci number divided by 2 to the n
fibonacci-numberssummation
fibonacci-numberssummation
Is there any way to calculate
$\sum_{n=1}^{\infty} \frac{F_{2n}}{2^{2n}}$
where $F_n$ is the nth Fibonacci number, $F_1=0$, $F_2=1$, $F_n = F_{n-1} + F_{n-2}$
Best Answer
The generating function of $F_{2n}$ (as defined here) is $$\frac{x-x^2}{x^2-3x+1}=\sum_{n=1}^\infty F_{2n}x^n$$ Substitute $x=\frac14$ to get the sum as $$\sum_{n=1}^\infty\frac{F_{2n}}{2^{2n}}=\frac{1/4-1/16}{1/16-3/4+1}=\frac35$$