Mathematical Reflections 2017, Issue 4 - Problem S415

Problem:
Let $$f(x)=\dfrac{(2x-1)6^x}{2^{2x-1}+3^{2x-1}}.$$
Evaluate $$f\left(\dfrac{1}{2018}\right)+f\left(\dfrac{3}{2018}\right)+\ldots+f\left(\dfrac{2017}{2018}\right).$$

Proposed by Titu Andreescu, University of Texas at Dallas, USA

Solution:
Observe that $$f(1-x)=\dfrac{(1-2x)6^{1-x}}{2^{1-2x}+3^{1-2x}}=\dfrac{(1-2x)6^x}{2^{2x-1}+3^{2x-1}}=-f(x),$$ i.e. $f(x)+f(1-x)=0$.
So,
\begin{eqnarray*}
\sum_{k=1}^{1009} f\left(\dfrac{2k-1}{2018}\right)&=&\sum_{k=1}^{504}\left(f\left(\dfrac{2k-1}{2018}\right)+f\left(\dfrac{2018-(2k-1)}{2018}\right)\right)+f\left(\dfrac{1009}{2018}\right)\\&=&f\left(\dfrac{1009}{2018}\right)\\&=&f\left(\dfrac{1}{2}\right)\\&=&0.
\end{eqnarray*}