If $f\left(x\right)=\frac{5^x}{5+5^x}$, then the value of $f\left(\frac{1}{20}\right)+f\left(\frac{2}{20}\right)+\dots +f\left(\frac{39}{20}\right)$ is

Solution:

We have,

$f\left(x\right)=\frac{5^x}{5+5^x}\Rightarrow f(2-x)=\frac{5^{2-x}}{5+5^{2-x}}=\frac{25}{5^{x+1}+5^2}=\frac{5}{5^x+5}$

$\therefore f\left(x\right)+f(2-x)=1$ for all x

Now,

$\begin{array}{l}f\left(\frac{1}{20}\right)+f\left(\frac{2}{20}\right)+\dots ..f\left(\frac{38}{20}\right)+f\left(\frac{39}{20}\right)\\ \begin{array}{r}=\left\{f\left(\frac{1}{20}\right)+f\left(\frac{39}{20}\right)\right\}+\left\{f\left(\frac{2}{20}\right)+f\left(\frac{38}{20}\right)\right\}+\dots \end{array}\\ +\left\{f\left(\frac{19}{20}\right)+f\left(\frac{21}{20}\right)\right\}+f\left(\frac{20}{20}\right)\\ =1\times 19+f\left(1\right)=19+\frac{5}{5+5}=\frac{39}{2}\end{array}$