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

We have $$f(x) = \frac{5^x}{5^x + 5}$$. Let us compute $$f(x) + f(2 - x)$$.

$$f(2 - x) = \frac{5^{2-x}}{5^{2-x} + 5} = \frac{25/5^x}{25/5^x + 5} = \frac{25}{25 + 5 \cdot 5^x} = \frac{5}{5 + 5^x}$$.

Therefore, $$f(x) + f(2 - x) = \frac{5^x}{5^x + 5} + \frac{5}{5 + 5^x} = \frac{5^x + 5}{5^x + 5} = 1$$.

The sum $$S = \sum_{k=1}^{39} f\left(\frac{k}{20}\right)$$ can be paired as $$f\left(\frac{k}{20}\right) + f\left(\frac{40-k}{20}\right) = 1$$ for $$k = 1, 2, \ldots, 19$$. This gives 19 pairs each summing to 1, plus the middle term $$f\left(\frac{20}{20}\right) = f(1) = \frac{5}{5 + 5} = \frac{1}{2}$$.

Therefore, $$S = 19 \cdot 1 + \frac{1}{2} = \frac{39}{2}$$.