2016 problem!

Relevant wiki: Function Problem Solving

Consider the following:

$\begin{aligned} f(x) + f(1-x) & = \frac 8{16^x+4} + \frac 8{16^{1-x}+4} \\ & = \frac 8{16^x+4} + \frac {8(16^x)}{16+4(16^x)} \\ & = \frac 8{16^x+4} + \frac {2(16^x)}{4+16^x} \\ & = \frac {8+2(16^x)}{16^x+4} \\ & = 2 \end{aligned}$

Now, let $x = \dfrac n{2016}$ , where $n$ is a positive integer, then $1-x = \dfrac {2016-n}{2016}$ . Therefore, the expression:

$\begin{aligned} \sum_{n=1}^{2016} f\left(\frac n{2016}\right) & = \sum_{n=1}^{1007} \left[ f\left(\frac n{2016}\right) + f\left(1 - \frac n{2016}\right) \right] + f\left(\frac {1008}{2016}\right) + f\left(\frac {2016}{2016}\right) \\ & = 2(1007) + \frac 8{16^\frac 12 +4} + \frac 8{16^1+4} \\ & = 2014 + 1 + 0.4 = \boxed{2015.400} \end{aligned}$

$f \left ( \dfrac x {2016}\right )+f \left (\dfrac{2016-x}{2016}\right )=8 \left \{ \dfrac 1 {16^{\frac x {2016}}+4}+\dfrac 1 {16^{\frac {2016 - x} {2016}}+4} \right \}=2.\\$ $\displaystyle \sum_{x=1}^{1007} \left \{ f \left ( \dfrac x {2016}\right )+f \left (\dfrac{2016-x}{2016}\right ) \right \}+ f \left ( \dfrac {1008} {2016} \right )+ f \left ( \dfrac {2016 }{2016}\right )\\$ $=2014+1+\frac 2 5=\huge \ \ \ \color{#D61F06}{2015.400}.\\$