TITLE: Sum of cubes of first n fibonacci numbers
QUESTION [10 upvotes]: 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 $\sum_{k=1}^n f_k^3$ ?

REPLY [12 votes]: $$F_n=\frac{1}{\sqrt{5}}(\phi^n-(-\phi)^{-n})\\
\sum_{k=1}^n F_k^3=\frac{1}{5\sqrt{5}}\sum_{k=1}^n(\phi^{3k}-3(-\phi)^k+3\phi^{-k}-(-\phi)^{-3k})$$
and then use Geometric Series

REPLY [11 votes]: 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.

REPLY [3 votes]: 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.