Sum of Functional Values for Symmetric Exponential Function

To find the value of the given expression, we first analyze a functional identity for $f(x)$.

The function is given by: $f(x) = \frac{4^x}{4^x + 2}$

Let us compute $f(1 - x)$: $f(1 - x) = \frac{4^{1-x}}{4^{1-x} + 2}$

Multiplying the numerator and the denominator by $4^x$, we get: $f(1 - x) = \frac{4}{4 + 2 \cdot 4^x} = \frac{2}{2 + 4^x}$

Now, adding $f(x)$ and $f(1 - x)$: $f(x) + f(1 - x) = \frac{4^x}{4^x + 2} + \frac{2}{4^x + 2} = \frac{4^x + 2}{4^x + 2} = 1$

Thus, for any $x \in [0, 1]$, we have the symmetry property: $f(x) + f(1 - x) = 1$

Now, consider the first part of the expression, which is the sum of 39 terms: $S_1 = f\left(\frac{1}{40}\right) + f\left(\frac{2}{40}\right) + f\left(\frac{3}{40}\right) + \dots + f\left(\frac{39}{40}\right)$

We can group the terms into pairs of the form $f\left(\frac{k}{40}\right) + f\left(\frac{40-k}{40}\right)$ for $k = 1, 2, \dots, 19$: $S_1 = \sum_{k=1}^{19} \left[ f\left(\frac{k}{40}\right) + f\left(1 - \frac{k}{40}\right) \right] + f\left(\frac{20}{40}\right)$

Using the identity $f(x) + f(1 - x) = 1$: $S_1 = \sum_{k=1}^{19} (1) + f\left(\frac{1}{2}\right) = 19 + f\left(\frac{1}{2}\right)$

The given expression requires us to find $S_1 - f\left(\frac{1}{2}\right)$: $\text{Value} = \left[ 19 + f\left(\frac{1}{2}\right) \right] - f\left(\frac{1}{2}\right) = 19$

Thus, the value of the expression is 19.