Sum of nth powers of Fibonacci numbers

Question

Is a closed form for $$\sum_{i=1}^n{F_i^k}$$ (where $F_i$ is the $i^{th}$ Fibonacci number and $k$ is constant) known?

Answer 1

According to Wikipedia, we have the following: $$\sum_{i=1}^n F_i=F_{n+2}-1$$ $$\sum_{i=1}^n F_i^2=F_nF_{n+1}$$

For sum of higher powers of Fibonacci numbers, look at this helpful blog post.

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One possible inuition behind these formulas is Binet's formula: $$F_n=\frac{1}{\sqrt 5}\left(\phi^n-\left(-\frac{1}{\phi}\right)^n\right)$$ Then, for any given constant $k$, we can use binomial expansion to get a bunch of terms that have the form $a\cdot b^n$ and we can find the sum of $\sum_{i=1}^n a\cdot b^i$ using the sum of geometric series formula. I'll apply this to the $k=1$ case: $$\sum_{i=1}^n \frac{1}{\sqrt 5}\left(\phi^i-\left(-\frac{1}{\phi}\right)^i\right)$$ Bring the $\frac{1}{\sqrt 5}$ out of the sumation: $$\frac{1}{\sqrt 5}\sum_{i=1}^n \left(\phi^i-\left(-\frac{1}{\phi}\right)^i\right)$$ Distribute the summation over the terms: $$\frac{1}{\sqrt 5}\left(\sum_{i=1}^n\phi^i-\sum_{i=1}^n\left(-\frac{1}{\phi}\right)^i\right)$$ Use geometric series formula: $$\frac{1}{\sqrt 5}\left(\phi\frac{\phi^n-1}{\phi-1}-\frac{1}{\phi}\frac{\left(-\frac{1}{\phi}\right)^n-1}{\frac{1}{\phi}+1}\right)$$ Use identities about $\phi$: $$\frac{1}{\sqrt 5}\left(\phi\frac{\phi^n-1}{\frac{1}{\phi}}-\frac{1}{\phi}\frac{\left(-\frac{1}{\phi}\right)^n-1}{\phi}\right)$$ Simplify: $$\frac{1}{\sqrt 5}\left(\phi^{n+2}-\phi^2-\left(-\frac{1}{\phi}\right)^{n+2}+\left(-\frac{1}{\phi}\right)^2\right)$$ Note that $\left(-\frac{1}{\phi}\right)^2-\phi^2=-\sqrt 5$: $$\frac{1}{\sqrt 5}\left(\phi^{n+2}-\left(-\frac{1}{\phi}\right)^{n+2}-\sqrt 5\right)$$ Partially distribute the $\frac{1}{\sqrt 5}$: $$\frac{1}{\sqrt 5}\left(\phi^{n+2}-\left(-\frac{1}{\phi}\right)^{n+2}\right)-\frac{1}{\sqrt 5}\cdot \sqrt 5$$ Use Binet's Formula and simplify: $$F_{n+2}-1$$ As you can see, this method is very cumbersome, so it's probably impractical for very high powers.