Let these two be summations of Fibonacci numbers:$$F_{2} + F_{5} + F_{8} + F_{11} + F_{14} +..... + F_{3n-1}=\sum\limits_{n=1} F_{3n-1}$$ and $$F_{0} + F_{3} + F_{6} + F_{9} + F_{12} +..... + F_{3n}=\sum\limits_{n=0} F_{3n}$$
Is there a closed form expression for these summations of Fibonacci numbers? I've taken a look at the Zeckendorf identities for these but I don't seem to get any useful progress in finding the formula.
On
Note : $$f_n=\frac{\phi^n-\psi^n}{\sqrt5} \text{ where } \phi=\frac{1+\sqrt5}{2}\text{ and } \psi=\frac{1-\sqrt5}{2}$$
Now $$\sum_{n\geq i\geq1}f_{3i-1}=\frac{1}{\sqrt5}\sum_{n\geq i\geq1}\phi^{3i-1}-\psi^{3i-1}\implies$$ $$\frac{1}{\sqrt5}\left(\frac{1}{\phi}\sum_{n\geq i\geq1}(\phi^3)^i-\frac{1}{\psi}\sum_{n\geq i\geq1}(\psi^3)^i\right)$$
Note : Sum of GP is $\frac{a(1-r^n)}{1-r}$ iff $|r|<1$
So , that gives you $$\frac{1}{\sqrt5}\left(\frac{\phi^2(1-\phi^{3n})}{1-\phi^3}-\frac{\psi^2(1-\psi^{3n})}{1-\psi^3}\right)$$
Similarly do for the other one , and then you can rationalize the denominators to get a bit more clean closed form for Summation of Fibonacci numbers .
$$\sum_{i=1}^nF_{3i-1}=\frac12(F_{3n+1}-1)$$ $$\sum_{i=0}^nF_{3i}=\frac12(F_{3n+2}-1)$$
You can deduce these simply by writing out a few terms; and you can prove them by induction.