# Sum of cubes of first n fibonacci numbers

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 $\sum_{k=1}^n f_k^3$ ?

$$F_n=\frac{1}{\sqrt{5}}(\phi^n-(-\phi)^{-n})\\ \sum_{k=1}^n F_k^3=\frac{1}{5\sqrt{5}}\sum_{k=1}^n(\phi^{3k}-3(-\phi)^k+3\phi^{-k}-(-\phi)^{-3k})$$
and then use Geometric Series

• It should be $F_n=\frac{1}{\sqrt{5}}(\phi^n-(-\phi)^{-n})$. – joriki May 6 '16 at 11:27

Any time you have a question like this, you should start by computing the first few terms of the sequence and then checking with the OEIS. In this case, the cubes of $1,1,2,3,5$ are $1,1,8,27,125$, leading to $1,2,10,37,162$, which is A005968. An answer to the question can be found there.

• The answer to the question should also be found right here, or else this is a link-only answer. – Chris Hayes Dec 5 '14 at 20:35
• @ChrisHayes, in this case I disagree. I gave quite a bit more than a link. At the same time, I was disinclined to give an utterly complete answer because the OP hadn't shown his or her own attempt to solve the problem or indicated where he/she was stuck. Also, if the OEIS disappears from online, then we're all sunk anyway ;-) – Barry Cipra Dec 5 '14 at 20:57

We have the identity $$\sum_{k=1}^n f_k^3=\frac{f_{3n+4}+6(-1)^nf_{n-1}+5}{10},$$ for all $n\ge 0$. For details see Recounting the Sums of Cubes of Fibonacci Numbers.

You could try playing around with a couple of Cassini's identities (as recurrence relations here):

\begin{eqnarray} F_{3n+1} &=& F_{n+1}^3 + 3 F_{n+1}F_n^2 - F_n^3 \\ F_{3n+2} &=& F_{n+1}^3 + 3 F_{n+1}^2F_n + F_n^3 \\ \end{eqnarray}