$f(\frac1{1234}) + f(\frac3{1234}) + f(\frac5{1234}) + ..... + f(\frac{1231}{1234}) + f(\frac{1233}{1234}) = ?$

Question

If $f(x) = \frac{e^{2x-1}}{1+e^{2x-1}}$, then how to evaluate $$f(\frac1{1234}) + f(\frac3{1234}) + f(\frac5{1234}) + ..... + f(\frac{1231}{1234}) + f(\frac{1233}{1234}) ?$$

I know I'm missing some trick here, what is it?

Answer 1

Using the trick you've identified,

\begin{align*} f(x) + f(1 - x) &= \frac{e^{2x - 1}}{1 + e^{2x - 1}} + \frac{e^{2(1 - x) - 1}}{1 + e^{2(1 - x) - 1}} \\ &= \frac{e^{2x - 1}(1 + e^{1 - 2x}) + e^{1 - 2x}(1 + e^{2x - 1})}{(1 + e^{2x - 1})(1 + e^{1 - 2x})} \\ &= \frac{e^{2x - 1} + 1 + e^{1 - 2x} + 1}{1 + e^{2x - 1} + e^{1 - 2x} + 1} \\ &= 1 \end{align*}

Now count how many terms there are.