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

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.
