A function $f\left( x \right)$ is given by $f\left( x \right) = \dfrac{{{5^x}}}{{{5^x} + 5}}$. Find the sum of the series $f\left( {\dfrac{1}{{20}}} \right) + f\left( {\dfrac{2}{{20}}} \right) + f\left( {\dfrac{3}{{20}}} \right) + \cdots + f\left( {\dfrac{{39}}{{20}}} \right)$.
A. $\dfrac{{19}}{2}$
B. $\dfrac{{49}}{2}$
C. $\dfrac{{39}}{2}$
D. $\dfrac{{29}}{2}$

Hint: First we will put $x = 2 - x$ in the given function. Then we will combine $f\left( {\dfrac{1}{{20}}} \right) + f\left( {\dfrac{{39}}{{20}}} \right)$, $f\left( {\dfrac{2}{{20}}} \right) + f\left( {\dfrac{{38}}{{20}}} \right)$, and put the value of each sum to calculate the value of given series.

Complete step by step solution:
Given series is $f\left( {\dfrac{1}{{20}}} \right) + f\left( {\dfrac{2}{{20}}} \right) + f\left( {\dfrac{3}{{20}}} \right) + \cdots + f\left( {\dfrac{{39}}{{20}}} \right)$.
Now putting $x = 2 - x$ in the function $f\left( x \right) = \dfrac{{{5^x}}}{{{5^x} + 5}}$
$f\left( {2 - x} \right) = \dfrac{{{5^{2 - x}}}}{{{5^{2 - x}} + 5}}$
               $ = \dfrac{{\dfrac{{{5^2}}}{{{5^x}}}}}{{\dfrac{{{5^2}}}{{{5^x}}} + 5}}$
               $ = \dfrac{{{5^2}}}{{{5^2} + 5 \cdot {5^x}}}$
Taking common 5 from the denominator and numerator
         $ = \dfrac{{5 \cdot 5}}{{5\left( {5 + {5^x}} \right)}}$
Cancel out 5 from the denominator and numerator
       $ = \dfrac{5}{{5 + {5^x}}}$
So, $f\left( x \right) + f\left( {2 - x} \right) = \dfrac{{{5^x}}}{{{5^x} + 5}} + \dfrac{5}{{{5^x} + 5}}$
                                        $ = \dfrac{{{5^x} + 5}}{{{5^x} + 5}}$
                                        $ = 1$
Rewrite the given series $f\left( {\dfrac{1}{{20}}} \right) + f\left( {\dfrac{2}{{20}}} \right) + f\left( {\dfrac{3}{{20}}} \right) + \cdots + f\left( {\dfrac{{39}}{{20}}} \right)$
$f\left( {\dfrac{1}{{20}}} \right) + f\left( {\dfrac{2}{{20}}} \right) + f\left( {\dfrac{3}{{20}}} \right) + \cdots + f\left( {\dfrac{{39}}{{20}}} \right)$
$ = \left[ {f\left( {\dfrac{1}{{20}}} \right) + f\left( {\dfrac{{39}}{{20}}} \right)} \right] + \left[ {f\left( {\dfrac{2}{{20}}} \right) + f\left( {\dfrac{{38}}{{20}}} \right)} \right] + \left[ {f\left( {\dfrac{3}{{20}}} \right) + f\left( {\dfrac{{37}}{{20}}} \right)} \right] + \cdots + \left[ {f\left( {\dfrac{{19}}{{20}}} \right) + f\left( {\dfrac{{21}}{{20}}} \right)} \right] + f\left( {\dfrac{{20}}{{20}}} \right)$
$ = \left[ {f\left( {\dfrac{1}{{20}}} \right) + f\left( {2 - \dfrac{1}{{20}}} \right)} \right] + \left[ {f\left( {\dfrac{2}{{20}}} \right) + f\left( {2 - \dfrac{2}{{20}}} \right)} \right] + \left[ {f\left( {\dfrac{3}{{20}}} \right) + f\left( {2 - \dfrac{3}{{20}}} \right)} \right] + \cdots + \left[ {f\left( {\dfrac{{19}}{{20}}} \right) + f\left( {2 - \dfrac{{19}}{{20}}} \right)} \right] + f\left( {\dfrac{{20}}{{20}}} \right)$$ = 1 + 1 + 1 + \cdots + 1 + f\left( 1 \right)$
Now putting $x = 1$ in $f\left( x \right) = \dfrac{{{5^x}}}{{{5^x} + 5}}$
$ \Rightarrow 1 + 1 + 1 + \cdots + 1 + \dfrac{{{5^1}}}{{{5^1} + 5}}$
$ \Rightarrow 1 + 1 + 1 + \cdots + 1 + \dfrac{5}{{2 \cdot 5}}$
$ \Rightarrow 1 + 1 + 1 + \cdots + 1 + \dfrac{1}{2}$
$ \Rightarrow 19 + \dfrac{1}{2}$
$ \Rightarrow \dfrac{{39}}{2}$

Option ‘B’ is correct

Note: We often tend to calculate the value of $1 + 1 + 1 + \cdots + 1 + f\left( 1 \right)$ and count as 39’s 1. But we make 19 groups and each group contains 2 functions. So there are 19’s 1.