$\text{ Let the function }f:[0,1]\to \mathrm{ℝ}\text{ be defined by }\mathrm{f}\left(\mathrm{x}\right)=\frac{4^{\mathrm{x}}}{4^{\mathrm{x}}+2}$

$\text{ Then the value of }\mathrm{f}\left(\frac{1}{40}\right)+\mathrm{f}\left(\frac{2}{40}\right)+\mathrm{f}\left(\frac{3}{40}\right)+\dots \dots \dots .+\mathrm{f}\left(\frac{39}{40}\right)-\mathrm{f}\left(\frac{1}{2}\right)\text{ is }$

$\begin{array}{l}f\left(x\right)+f(1-x)=\frac{4^x}{4^{\mathrm{x}}+2}+\frac{4^{1-x}}{4^{1-x}+2}\\ =\frac{4^x}{4^x+2}+\frac{4/4^x}{\frac{4}{4^x}+2}\\ =\frac{4^x}{4^x+2}+\frac{4}{4+2\cdot 4^x}\\ =\frac{4^x}{4^x+2}+\frac{2}{2+4^{\mathrm{x}}}\\ =1\\ \text{ so, }f\left(\frac{1}{40}\right)+f\left(\frac{2}{40}\right)+\dots +f\left(\frac{39}{40}\right)-f\left(\frac{1}{2}\right)\\ =19+f\left(\frac{1}{2}\right)-f\left(\frac{1}{2}\right)=19\end{array}$