 fluorekyz4l

2023-02-26

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$\frac{29}{2}$ Gustavo Gamble

The ideal decision is C $\frac{39}{2}$
Step 1: Finding $f\left(2-x\right)$
The function is $f\left(x\right)=\frac{{5}^{x}}{{5}^{x}+5}$.
Finding the sum requires finding $f\left(2-x\right)$
$f\left(2-x\right)=\frac{{5}^{2-x}}{{5}^{2-x}+5}=\frac{\frac{{5}^{2}}{{5}^{x}}}{\frac{{5}^{2}}{{5}^{x}}+5}=\frac{5}{5+{5}^{x}}$
Step 2: Add $f\left(x\right),f\left(2-x\right)$
$f\left(x\right)+f\left(2-x\right)=\frac{{5}^{x}}{{5}^{x}+5}+\frac{5}{{5}^{x}+5}=1$
Now consider, $x=\frac{1}{20}$
$⇒f\left(x\right)=f\left(\frac{1}{20}\right)⇒f\left(2-x\right)=f\left(2-\frac{1}{20}\right)=f\left(\frac{39}{20}\right)$
Step 3:Finding the pairs
$f\left(\frac{1}{20}\right)+f\left(\frac{39}{20}\right)=1\left(\because f\left(x\right)+f\left(2-x\right)=1\right)$
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
$⇒\left(19\right)\left(1\right)+\frac{1}{2}⇒\frac{39}{2}$
Hence, option (C) is the correct answer.

Do you have a similar question?