Key Concepts and Formulas

• Functional Symmetry: A function $f(x)$ might exhibit symmetry properties, such as $f(x) + f(a-x) = C$ for some constant $C$. This property is invaluable for simplifying sums of series by pairing terms.
• Exponent Rules: Understanding rules like $a^{m-n} = a^m / a^n$ is crucial for algebraic manipulation of exponential functions.
• Arithmetic Series Summation: For a series with an odd number of terms, there is a middle term. When pairing terms from the ends that sum to a constant, the total sum is the sum of the pairs plus the middle term.

Step-by-Step Solution

Step 1: Discovering a Key Functional Property We are given the function $f(x) = \frac{5^x}{5^x + 5}$. Let's investigate the property $f(x) + f(a-x)$ for a suitable constant $a$. The arguments in the series are of the form $\frac{k}{20}$ for $k=1, 2, \dots, 39$. The sum of the first and last argument is $\frac{1}{20} + \frac{39}{20} = \frac{40}{20} = 2$. This suggests we should check the property $f(x) + f(2-x)$.

First, find $f(2-x)$: $f(2-x) = \frac{5^{2-x}}{5^{2-x} + 5}$ Using the exponent rule $5^{2-x} = \frac{5^2}{5^x} = \frac{25}{5^x}$, we get: $f(2-x) = \frac{\frac{25}{5^x}}{\frac{25}{5^x} + 5}$ To simplify, multiply the numerator and denominator by $5^x$: $f(2-x) = \frac{25}{25 + 5 \cdot 5^x} = \frac{25}{5(5 + 5^x)} = \frac{5}{5 + 5^x}$ Now, let's add $f(x)$ and $f(2-x)$: $f(x) + f(2-x) = \frac{5^x}{5^x + 5} + \frac{5}{5^x + 5} = \frac{5^x + 5}{5^x + 5} = 1$ Thus, we have discovered the property $f(x) + f(2-x) = 1$.

Step 2: Applying the Property to the Series Sum The series is $S = f\left( \frac{1}{20} \right) + f\left( \frac{2}{20} \right) + \dots + f\left( \frac{39}{20} \right)$. There are 39 terms in the series. Since 39 is odd, there will be a middle term. The middle term is the $\frac{39+1}{2} = 20^{th}$ term, which is $f\left(\frac{20}{20}\right) = f(1)$.

We can pair the terms using the property $f(x) + f(2-x) = 1$. The pairs are:

• $f\left(\frac{1}{20}\right) + f\left(2 - \frac{1}{20}\right) = f\left(\frac{1}{20}\right) + f\left(\frac{39}{20}\right) = 1$
• $f\left(\frac{2}{20}\right) + f\left(2 - \frac{2}{20}\right) = f\left(\frac{2}{20}\right) + f\left(\frac{38}{20}\right) = 1$
• ...
• $f\left(\frac{19}{20}\right) + f\left(2 - \frac{19}{20}\right) = f\left(\frac{19}{20}\right) + f\left(\frac{21}{20}\right) = 1$

There are $\frac{39-1}{2} = 19$ such pairs, and each pair sums to 1. The sum of these 19 pairs is $19 \times 1 = 19$.

Step 3: Calculating the Middle Term and the Total Sum The middle term of the series is $f\left(\frac{20}{20}\right) = f(1)$. $f(1) = \frac{5^1}{5^1 + 5} = \frac{5}{5+5} = \frac{5}{10} = \frac{1}{2}$ The total sum of the series is the sum of the pairs plus the middle term: $S = (\text{Sum of 19 pairs}) + (\text{Middle term})$ $S = 19 + \frac{1}{2}$ To add these, we find a common denominator: $S = \frac{19 \times 2}{2} + \frac{1}{2} = \frac{38}{2} + \frac{1}{2} = \frac{39}{2}$

Common Mistakes & Tips

• Incorrect Pairing: Ensure that the arguments sum up to the correct constant (in this case, 2) when applying the symmetry property.
• Handling the Middle Term: For series with an odd number of terms, always identify and calculate the middle term separately after pairing.
• Algebraic Errors: Be meticulous with algebraic manipulations, especially when dealing with exponents and fractions.

Summary The problem involves summing a series of function values. By discovering the functional property $f(x) + f(2-x) = 1$, we were able to pair terms from the beginning and end of the series. Each pair sums to 1. With 39 terms, there are 19 such pairs and one middle term, $f(1)$. The sum of the pairs is 19, and the middle term is $1/2$. Therefore, the total sum of the series is $19 + 1/2 = 39/2$.

The final answer is \boxed{\frac{39}{2}}.