2015 and Fibonacci

Let $a_1$ , $a_2$ , $a_3$ , $\dots$ , $a_{2014}$ , and $a_{2015}$ be real numbers and also let

$f(x) = \begin{cases}x + a_1 & & x \le 1 \\ 2x + a_2 & & 1 < x \le 2 \\ 3x + 2a_3 & & 2 < x \le 3 \\ \quad \cdots & & \quad \cdots \\ 2014x + F_{2014}a_{2014} & & 2013 < x \le 2014 \\2015x + F_{2015}a_{2015} & & 2014 < x \le 2015 \\ 2016x + F_{2016}a_{1} & & x > 2015 \end{cases}$

where $F_n$ is the $n$ th Fibonacci number .

If $f\left( x \right)$ is continuous everywhere, then the maximum value of $a_1 + a_2 + a_3 + \dots + a_{2014} + a_{2015}$ can be expressed as $-\dfrac { \alpha }{ F_\beta - F_\gamma - 1 }$ , where $\alpha$ , $\beta$ , and $\gamma$ are positive integers. Find the value of $\alpha + \beta + \gamma$ .

Please show your solutions,

Image credit: Flickr Jean