Hidden Property Trick of A Specific Function

The trick here is that for $a \in \mathbb{R}$ , we have $f(a) + f(1-a) = 1$ . Let's prove this:

Let $a \in \mathbb{R}$ . Then, $f(a)+f(1-a) = \frac{2}{2+4^a} + \frac{2}{2+4^{1-a}} = \frac{2(2+4^{1-a})}{(2+4^a)(2+4^{1-a})} + \frac{2(2+4^a)}{(2+4^{1-a})(2+4^a)} = \frac{8+2*4^{1-a}+2*4^a}{8+2*4^{1-a}+2*4^a} = 1$

Therefore, $f(\frac{1}{2014}) + f(\frac{2013}{2014}) = 1$ , $f(\frac{2}{2014}) + f(\frac{2012}{2014}) = 1$ , and so on. There are 1006 such pairs which together add to 1006. Then, the remaining term is \(f(\frac{1007}{2014}). This can be simplified:

\(f(\frac{1007}{2014}) = f(\frac{1}{2}) = \frac{2}{2+4^{1/2}} = \frac{1}{2}\)

Hence, the total sum is 1006.5.