Let $\{f_k\}$ be the sequence of fibonacci numbers. It is well-known that $\sum_{k=1}^n f_k=f_{n+2}-1$ and $\sum_{k=1}^n f_k^2=f_n f_{n+1}$ . Is there a formula for $\sum_{k=1}^n f_k^3$ ?
[Math] Sum of cubes of first n fibonacci numbers
fibonacci-numberssequences-and-series
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From the standard theory of linear recurrence,s $F_n$ is the positive real root of the equation $z^n = z^{n-1} + z^{n-2} + \cdots + z + 1$. Multiplying both sides by $1-z$ and rearranging, you get that $F_n$ is the positive real root of $f_n(z) = z^{n+1} - 2z^n + 1 = 0$ that is not $z = 1$. (By Descartes' rule of signs (https://en.wikipedia.org/wiki/Descartes%27_rule_of_signs), there are either 0 or 2 positive real roots of this polynomial; we know there is at least 1 , so there must be 2.)
Now, we have $$ f_n(2)= 2^{n+1} - 2 \cdot 2^{n} + 1 = 1 $$ and $$f_n(2 - 1/2^{n-1}) = (2 - 1/2^{n-1})^{n+1} - 2 \cdot (2-1/2^{n-1})^n + 1. $$ We aim to show that $f_n(2-1/2^{n-1}) < 0$; then $f_n$ must have a root between $2 - 1/2^{n-1}$ and $2$.
Factoring out powers of 2, we get $$f_n(2 - 1/2^{n-1}) = 2^{n+1} (1-1/2^n)^{n+1} - 2^{n+1} (1-1/2^n)^n + 1.$$ Factoring out $2^{n+1} (1-1/2^n)^n$ from the first two terms gives $$f_n(2 - 1/2^{n-1}) = 2^{n+1} (1-1/2^n)^n (-1/2^n) + 1$$ or, finally, $$f_n(2 - 1/2^{n-1}) = 1-2(1-1/2^n)^n. $$ So the result that $f_n$ has a root in the interval $[2-1/2^{n-1}, 2]$, for $n$ sufficiently large, follows from the fact that $(1-1/2^n)^n > 1/2$ for $n$ sufficiently large. Since $\lim_{n \to \infty} (1-1/2^n)^n = 1$ (use for example L'Hopital's rule) this follows.
Thus $F_n \in [2 - 1/2^{n-1}, 2]$ and the desired result follows, for example, from the squeeze theorem.
Short answer: No, because the Fibonacci sequence grows exponentially.
Longer answer: First suppose $\{n,m\} \cap \{k,\ell\} \neq \emptyset$, say $n=l$. Then $N = F_m - F_k$. This implies that $N = F_m - F_k \geq F_m - F_{m-1} = F_{m-2}$. It follows that $m = O(\log(N))$ because the Fibonacci sequence is exponential. Similarly we have $k = O(\log(N))$. It follows that for large $N$ we can choose $m$ and $k$ in at most $O(\log(N)^2)$ ways.
Now suppose $\{n,m\}$ and $\{k,\ell\}$ are disjoint. Then we similarly find that we can choose $n$, $m$, $k$, $\ell$ in at most $O(\log(N)^4)$ ways.
Consequently, to write the numbers $1$, $2$, ..., $N$ in the desired form, there are at most $O(\log(N)^2 + \log(N)^4)$ possible choices for $n$, $m$, $k$, $\ell$. However this becomes strictly smaller than $N$ for large $N$.
Best Answer
$$F_n=\frac{1}{\sqrt{5}}(\phi^n-(-\phi)^{-n})\\ \sum_{k=1}^n F_k^3=\frac{1}{5\sqrt{5}}\sum_{k=1}^n(\phi^{3k}-3(-\phi)^k+3\phi^{-k}-(-\phi)^{-3k})$$
and then use Geometric Series