If f(x)=5x5+5x, then the value of f120+f220+…+f3920 is

A
20
B
$\frac{29}{2}$
C
$\frac{19}{2}$
D
$\frac{39}{2}$

Solution:

We have,

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

$∴ f (x) + f (2 - x) = 1$ for all x

Now,

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

If $f (x) = \frac{5^{x}}{5 + 5^{x}}$ , then the value of $f (\frac{1}{20}) + f (\frac{2}{20}) + \dots + f (\frac{39}{20})$ is

Solution:

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