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

As suggested, the simplest would consist in a use of Binet formula for Fibonacci number. It write $$F_n=\frac{\varphi ^n-\psi ^n}{\sqrt{5}}$$ with $\varphi=\frac{1}{2} \left(1+\sqrt{5}\right)$ and  $\psi=\frac{1}{2} \left(1-\sqrt{5}\right)$. So, if we consider each term of the sum, we can rewrite it as $$F_{3n}2^{-3n}=\frac{\Phi ^n-\Psi ^n}{\sqrt{5}}$$ noting $$\Phi=\frac{1}{8} \left(1+\sqrt{5}\right)^3$$ and $$\Psi=\frac{1}{8} \left(1-\sqrt{5}\right)^3$$ Now, we consider the partial sum from $n=1$ to $n=m$ and apply the rules of geometric progressions. So we obtain $$\sum _{n=1}^m F_{3n}2^{-3n}=\frac{\frac{\Phi  \left(\Phi ^m-1\right)}{\Phi -1}-\frac{\Psi  \left(\Psi
   ^m-1\right)}{\Psi -1}}{\sqrt{5}}$$
