Let $f(x)=\frac{4^{x}}{4^{x}+2}$. Calculate the value of

$$
f(0 / 2015)+f(1 / 2015)+f(2 / 2015)+\ldots+f(2014 / 2015)+f(2015 / 2015)
$$
Solution. First, we show that

$$
\begin{gathered}
f\left(\frac{k}{n}\right)+f\left(\frac{n-k}{n}\right)=1, \quad \text { for all } n, k \in \mathbb{N}^{+}, 0 \leq k \leq n. \\
\frac{4^{\frac{k}{n}}}{4^{\frac{k}{n}}+2}+\frac{4^{\frac{n-k}{k}}}{4^{\frac{n-k}{n}}+2}=\frac{4^{\frac{k}{n}}}{4^{\frac{k}{n}}+2}+\frac{4^{1-\frac{k}{n}}}{4^{1-\frac{k}{n}}+2}=\frac{4^{\frac{k}{n}}}{4^{\frac{k}{n}}+2}+\frac{\frac{4}{4^{\frac{k}{n}}}}{\frac{4^{\frac{k}{n}}}{4^{n}}+2}= \\
=\frac{4^{\frac{k}{n}}}{4^{\frac{k}{n}}+2}+\frac{4}{4+2 \cdot 4^{\frac{k}{n}}}=\frac{4^{\frac{k}{n}}}{4^{\frac{k}{n}}+2}+\frac{2}{4^{\frac{k}{n}}+2}=1
\end{gathered}
$$

(We can prove that $f(x)+f(1-x)=1$ for any real $x$, but for the solution, the above consideration is sufficient.) To complete the solution, we form pairs from the 2016 function values:

$$
\begin{gathered}
f(0 / 2015)+f(2015 / 2015)=f(1 / 2015)+f(2014 / 2015)=\ldots= \\
=f(1007 / 2015)+f(1008 / 2015)=1 .
\end{gathered}
$$

Since the sum in each pair is 1, the total sum will be 1008.