TITLE: Sum of nth powers of Fibonacci numbers
QUESTION [6 upvotes]: 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?

REPLY [2 votes]: Consider a generalized Fibonacci sequence $\{U_n\}$ with $U_n=U_n(p,q)$ defined as $U_0=0$, $U_1=1$, and $U_{n+2}=-qU_n+pU_{n+1}$ for all $n\geq0$. It is prove that
$$\sum_{i=1}^nU_i^m=\frac{1}{\sum_{i=1}^m{m\atopwithdelims\{\}i}}\left(\sum_{i=1}^nU_{im-\binom{m+1}{2}}+\sum_{i=1}^m{m\atopwithdelims\{\}i}\sum_{j=1}^i(U_{n-i+j}^m-U_{j-i}^m)\right),$$
in which 
$${m\atopwithdelims\{\}i}=\left(\prod_{\underset{j=1}{j\neq i}}^mU_{j-i}\right)^{-1}.$$
Also, note that in the above formula
$$\sum_{x=1}^nU_{ax+b}=\frac{q^aU_{na+b}-U_{(n+1)a+b}-q^aU_{b-a}+U_b}{1+q^a-V_a}-U_b$$
for all integers $a,b$. See here.