A function f(x) is given by f(x)=5x5x+5, then the sum of the series: f120+f220+f320+......+f3920 is...

The ideal decision is C $\frac{39}{2}$
Explanation for the correct answer:
Step 1: Finding $f(2-x)$
The function is $f\left(x\right)=\frac{5^x}{5^x+5}$.
Finding the sum requires finding $f(2-x)$
$f(2-x)=\frac{5^{2-x}}{5^{2-x}+5}=\frac{\frac{5^2}{5^x}}{\frac{5^2}{5^x}+5}=\frac{5}{5+5^x}$
Step 2: Add $f\left(x\right),f(2-x)$
$f\left(x\right)+f(2-x)=\frac{5^x}{5^x+5}+\frac{5}{5^x+5}=1$
Now consider, $x=\frac{1}{20}$
$\Rightarrow f\left(x\right)=f\left(\frac{1}{20}\right)\Rightarrow f(2-x)=f\left(2-\frac{1}{20}\right)=f\left(\frac{39}{20}\right)$
Step 3:Finding the pairs
$f\left(\frac{1}{20}\right)+f\left(\frac{39}{20}\right)=1\left(\because f\left(x\right)+f(2-x)=1\right)$
Similarly, the pairs are
$f\left(\frac{2}{20}\right)+f\left(\frac{38}{20}\right)=1,....................f\left(\frac{19}{20}\right)+f\left(\frac{21}{20}\right)=1$
and
$f\left(\frac{20}{20}\right)=f\left(1\right)f\left(1\right)=\frac{5^1}{5^1+5}=\frac{1}{2}$
Step 4: Finding the summation
$f\left(\frac{1}{20}\right)+f\left(\frac{2}{20}\right)+f\left(\frac{3}{20}\right)+......+f\left(\frac{39}{20}\right)$
There are $19$ such pairs and $f\left(1\right)$
Consequently, the series' total is
$\Rightarrow \left(19\right)\left(1\right)+\frac{1}{2}\Rightarrow \frac{39}{2}$
Hence, option (C) is the correct answer.