If f(x)=5×5+5x, then the value of f120+f220+…+f3920 is

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

1. A

20

2. B

$\frac{29}{2}$

3. C

$\frac{19}{2}$

4. D

$\frac{39}{2}$

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Solution:

We have,

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

for all x

Now,

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

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