Step-by-step Solution:

1. Analyze the given function. The function is given by $f(x) = \frac{4^x}{4^x + 2}$, defined on the interval $[0, 1]$. We need to evaluate the expression: $E = f\left( {{1 \over {40}}} \right) + f\left( {{2 \over {40}}} \right) + f\left( {{3 \over {40}}} \right) + ... + f\left( {{{39} \over {40}}} \right) - f\left( {{1 \over 2}} \right)$

2. Investigate a key property of the function. Let's examine the sum $f(x) + f(1-x)$. This is a common strategy for sums with symmetric arguments. $f(1-x) = \frac{4^{1-x}}{4^{1-x} + 2}$ To simplify this, we can write $4^{1-x}$ as $\frac{4}{4^x}$: $f(1-x) = \frac{\frac{4}{4^x}}{\frac{4}{4^x} + 2}$ Multiply the numerator and the denominator by $4^x$: $f(1-x) = \frac{4}{4 + 2 \cdot 4^x} = \frac{2 \cdot 2}{2(2 + 4^x)} = \frac{2}{4^x + 2}$ Now, let's compute the sum $f(x) + f(1-x)$: $f(x) + f(1-x) = \frac{4^x}{4^x + 2} + \frac{2}{4^x + 2} = \frac{4^x + 2}{4^x + 2} = 1$ So, we have the very useful property: $f(x) + f(1-x) = 1$.

3. Apply the property to the sum in the expression. Let $S = f\left( {{1 \over {40}}} \right) + f\left( {{2 \over {40}}} \right) + ... + f\left( {{{39} \over {40}}} \right)$. This is a sum of 39 terms. We can pair the terms in the sum. The first term is paired with the last, the second with the second-to-last, and so on. $S = \left[ f\left(\frac{1}{40}\right) + f\left(\frac{39}{40}\right) \right] + \left[ f\left(\frac{2}{40}\right) + f\left(\frac{38}{40}\right) \right] + \dots$ Let's consider a general pair: $f\left(\frac{k}{40}\right) + f\left(\frac{40-k}{40}\right)$. Since $\frac{40-k}{40} = 1 - \frac{k}{40}$, this pair is of the form $f(x) + f(1-x)$ with $x = \frac{k}{40}$. From our property, each such pair sums to 1. $f\left(\frac{k}{40}\right) + f\left(1-\frac{k}{40}\right) = 1$

4. Evaluate the sum S. The sum has terms from $k=1$ to $k=39$. We can form pairs for $k=1, 2, \dots, 19$.

   • For $k=1$: $f(\frac{1}{40}) + f(\frac{39}{40}) = 1$
   • For $k=2$: $f(\frac{2}{40}) + f(\frac{38}{40}) = 1$
   • ...
   • For $k=19$: $f(\frac{19}{40}) + f(\frac{21}{40}) = 1$ There are 19 such pairs. The sum of these pairs is $19 \times 1 = 19$. The middle term of the series, when $k=20$, is $f\left(\frac{20}{40}\right) = f\left(\frac{1}{2}\right)$, which is left unpaired. So, the total sum is: $S = (\text{sum of 19 pairs}) + (\text{middle term})$ $S = 19 + f\left(\frac{1}{2}\right)$

5. Calculate the final expression. The expression we need to find is $E = S - f\left(\frac{1}{2}\right)$. Substituting the value of S we found: $E = \left( 19 + f\left(\frac{1}{2}\right) \right) - f\left(\frac{1}{2}\right)$ $E = 19$

Alternative Method for the sum: Let $S = \sum_{k=1}^{39} f\left(\frac{k}{40}\right)$. We can also write the sum in reverse order: $S = \sum_{k=1}^{39} f\left(\frac{40-k}{40}\right) = \sum_{k=1}^{39} f\left(1-\frac{k}{40}\right)$. Adding the two expressions for S: $2S = \sum_{k=1}^{39} f\left(\frac{k}{40}\right) + \sum_{k=1}^{39} f\left(1-\frac{k}{40}\right) = \sum_{k=1}^{39} \left[ f\left(\frac{k}{40}\right) + f\left(1-\frac{k}{40}\right) \right]$ Using the property $f(x)+f(1-x)=1$, we get: $2S = \sum_{k=1}^{39} 1 = 39$. So, $S = \frac{39}{2}$. The required expression is $E = S - f\left(\frac{1}{2}\right)$. Let's calculate $f(\frac{1}{2})$: $f(\frac{1}{2}) = \frac{4^{1/2}}{4^{1/2}+2} = \frac{2}{2+2} = \frac{1}{2}$. $E = \frac{39}{2} - \frac{1}{2} = \frac{38}{2} = 19$.

Both methods yield the same result.