Compute the sum $$\sum_{a_{2015} = 0}^{\infty} \sum_{a_{2014} = 0}^{a_{2015}} \sum_{a_{2013} = 0}^{a_{2014}} \cdots \sum_{a_{1} = 0}^{a_2} \sum_{k=0}^{a_1} \frac{F_{k}}{2^{a_{2015}}} $$ where $F_k$ denotes the $k$th Fibonacci number ($F_0 = 1$, $F_1 = 1$, $F_{n} = F_{n-1} + F_{n-2}$).
I have no idea how to begin! Any help would be much appreciated.
It is well known that $1/(1-x-x^2)$ is the generating function for the Fibonacci numbers. Observe that $$c_n = \sum_{a_{2014}=0}^{n} \sum_{a_{2013} = 0}^{a_{2014}} \cdots \sum_{k=0}^{a_1} F_k$$ is the coefficient of $x^n$ in the power series expansion of $$f(x) = \frac{1}{(1-x)^{2015} (1-x-x^2)}$$ The desired sum is $$\sum_{n=0}^{\infty} \frac{c_n}{2^n} = f(1/2) = 2^{2017}$$