partial Fibonacci summation Let $F_{n}$ be the n-th Fibonacci number.
How to calculate the summation like following:
$\sum_{n \geq 0}  F_{3n} \cdot 2^{-3n}$

Here's an approach via generating functions. As the Fibonacci recurrence is defined by $F_{n+2} = F_{n+1} + F_n$, we have
$$\sum_{n \ge 0} F_{n+2}z^{n+2} = \sum_{n \ge 0} F_{n+1}z^{n+1}z + \sum_{n \ge 0}F_nz^nz^2$$
which with the generating function $G(z) = \sum_{n\ge0} F_n z^n$ gives
$$G(z) - F_0 - F_1z = zG(z) - zF_0 + z^2G(z)$$
and therefore (using $F_0 = 0$ and $F_1 = 1$),
$$G(z) - z = zG(z) + z^2G(z) \implies G(z) = \frac{z}{1 - z - z^2}.$$
This much is well-known. Now let $\omega$ be a third root of unity, so that $\omega^3 = 1$. Then 
$$G(z) + G(z\omega) + G(z\omega^2) = \sum_{n\ge0} F_nz^n(1 + \omega^n + \omega^{2n}) = \sum_{n\ge0} 3F_{3n}z^{3n},$$
as we have
$$1 + \omega^n + \omega^{2n} = \begin{cases} 3 \text{ if $3$ divides $n$}\\0 \text{ otherwise.}\end{cases}$$
This means that the number $\sum_{n\ge0} F_{3n}2^{-3n}$ we want is
$$\frac{G(z) + G(z\omega) + G(z\omega^2)}{3}$$
with $z = \frac12$. The sum turns out to be
$$\frac13\left(\frac{1/2}{1-1/2-(1/2)^2} + \frac{\omega(1/2)}{1-\omega(1/2)-\omega^2(1/2)^2} + \frac{\omega^2(1/2)}{1-\omega^2(1/2)-\omega(1/2)^2}\right)$$
$$=\frac13\left(2 - \frac{14}{31}\right) = \frac{16}{31}.$$
