Question

A function $f\left(x\right)$ is given by $f\left(x\right)=\frac{5^x}{5^x+5}$, then the sum of the series: $f\left(\frac{1}{20}\right)+f\left(\frac{2}{20}\right)+f\left(\frac{3}{20}\right)+......+f\left(\frac{39}{20}\right)$ is equal to:

• A. $\frac{19}{2}$
• B. $\frac{49}{2}$
• C. $\frac{39}{2}$
• D. $\frac{29}{2}$

Answer 1

The correct option is C

$\frac{39}{2}$

Explanation for the correct answer:

Step 1: Finding $f(2-x)$

The function is $f\left(x\right)=\frac{5^x}{5^x+5}$.

To find the sum, we need to find $f(2-x)$

$$\begin{gathered} f(2-x)=\frac{5^{2-x}}{5^{2-x}+5} \\ =\frac{{\displaystyle \frac{5^2}{5^x}}}{{\displaystyle \frac{5^2}{5^x}}+5} \\ =\frac{5}{5+5^x} \end{gathered}$$

Step 2: Add $f\left(x\right),f(2-x)$

$$\begin{gathered} f\left(x\right)+f(2-x)=\frac{5^x}{5^x+5}+\frac{5}{5^x+5} \\ =1 \end{gathered}$$

Now consider, $x=\frac{1}{20}$

$$\begin{gathered} \Rightarrow f\left(x\right)=f\left(\frac{1}{20}\right) \\ \Rightarrow f(2-x)=f\left(2-\frac{1}{20}\right) \\ =f\left(\frac{39}{20}\right) \end{gathered}$$

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

$$\begin{gathered} f\left(\frac{2}{20}\right)+f\left(\frac{38}{20}\right)=1, \\ ..... \\ ..... \\ ..... \\ ..... \\ f\left(\frac{19}{20}\right)+f\left(\frac{21}{20}\right)=1 \end{gathered}$$

and

$$\begin{gathered} f\left(\frac{20}{20}\right)=f\left(1\right) \\ f\left(1\right)=\frac{5^1}{5^1+5} \\ =\frac{1}{2} \end{gathered}$$

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 $f\left(1\right)$

Thus, the sum of the series is

$$\begin{gathered} \Rightarrow \left(19\right)\left(1\right)+\frac{1}{2} \\ \Rightarrow \frac{39}{2} \end{gathered}$$

Hence, option (C) is the correct answer.