Proving $\sum_{k=0}^{\infty} F_{mk}z^k=\frac{F_mz}{1-z(F_{m-1}+F_{m+1})+(-1)^mz^2}$

Question

I have read in a few places that$$\sum_{k=0}^{\infty} F_{mk}z^k=\frac{F_mz}{1-z(F_{m-1}+F_{m+1})+(-1)^mz^2}$$where $F_i$ denotes the $i$-th Fibonacci number. The series is a generalization of the more known $$\sum_{k=0}^{\infty} F_{k}z^k=\frac{1}{1-z-z^2}$$The latter is relatively easy to prove, as done here, but I havent been able find a proof of the former.

Answer 1

Try using Binet’s formula for the Fibonacci numbers: $$F_n=\frac{1}{\sqrt{5}}\bigg(\frac{1+\sqrt{5}}{2}\bigg)^n-\frac{1}{\sqrt{5}}\bigg(\frac{1-\sqrt{5}}{2}\bigg)^n$$ Combining this with the formula for the sum of a geometric series, we have that your sum is equal to $$\begin{align} \sum_{k=0}^\infty F_{mk}z^k &= \frac{1}{\sqrt{5}}\sum_{k=0}^\infty (\phi^{mk}-(-\phi)^{-mk})z^k \\ &= \frac{1}{\sqrt{5}}\bigg(\frac{1}{1-\phi^m z} - \frac{1}{1-(-\phi)^{-m}z}\bigg) \\ &= \frac{1}{\sqrt{5}}\frac{1-(-\phi)^{-m} z-1+\phi^{m}z}{(1-\phi^m z)(1-(-\phi)^{-m}z)} \\ &= \frac{1}{\sqrt{5}} \frac{\phi^m z-(-\phi)^{-m}z}{1-(\phi^m+(-\phi)^{-m})z+(-1)^m z^2} \\ &= \frac{F_m z}{1-(F_{m-1}+F_{m+1})z+(-1)^m z^2} \end{align}$$