Question

Let the function $f:[0,1]\to R$ be defined by
$f(x)=\frac{4^x}{4^x+2}$
Then the value of
$f\left(\frac{1}{40}\right)+f\left(\frac{2}{40}\right)+f\left(\frac{3}{40}\right)+\dots +f\left(\frac{39}{40}\right)-f\left(\frac{1}{2}\right)$ is ____

Answer 1

Answer: 19.00

$\sum _{r=1}^{39}f\left(\frac{r}{40}\right)=\frac{1}{2}\sum _{r=1}^{39}f\left(\frac{r}{40}\right)+f\left(\frac{40-r}{40}\right)=\frac{39}{2}$
As $(f(x)+f(1-x)=1)\&f\left(\frac{1}{2}\right)=\frac{1}{2}$
So $\sum _{r=1}^{39}f\left(\frac{r}{40}\right)-f\left(\frac{1}{2}\right)=\frac{39}{2}-\frac{1}{2}=19$