Key Concepts and Formulas

• Functional Symmetry: Identifying and exploiting properties of a function where $f(x) + f(a-x)$ or $f(x) + f(a+x)$ equals a constant is crucial for simplifying sums of function values.
• Arithmetic Progression: Recognizing that the arguments of the function form an arithmetic progression can help in pairing terms.
• Summation of Series: Efficiently summing series by pairing terms and accounting for any middle term.

Step-by-Step Solution

Step 1: Analyze the Function and the Sum We are given the function $f(x) = \frac{2e^{2x}}{e^{2x} + e}$ and asked to compute the sum $S = f\left( \frac{1}{100} \right) + f\left( \frac{2}{100} \right) + \dots + f\left( \frac{99}{100} \right)$. The arguments of the function form an arithmetic progression $1/100, 2/100, \dots, 99/100$. Notice that the sum of the first and last argument is $1/100 + 99/100 = 1$, the sum of the second and second-to-last argument is $2/100 + 98/100 = 1$, and so on. This suggests investigating the property $f(x) + f(1-x)$.

Step 2: Derive the Functional Property $f(x) + f(1-x)$ We need to find the value of $f(x) + f(1-x)$. First, let's find an expression for $f(1-x)$: $f(1-x) = \frac{2e^{2(1-x)}}{e^{2(1-x)} + e}$ Simplify the term $e^{2(1-x)}$: $e^{2(1-x)} = e^{2-2x} = e^2 \cdot e^{-2x} = \frac{e^2}{e^{2x}}$ Substitute this back into the expression for $f(1-x)$: $f(1-x) = \frac{2 \left( \frac{e^2}{e^{2x}} \right)}{\left( \frac{e^2}{e^{2x}} \right) + e}$ To simplify the complex fraction, multiply the numerator and denominator by $e^{2x}$: $f(1-x) = \frac{2 \left( \frac{e^2}{e^{2x}} \right) \cdot e^{2x}}{\left( \frac{e^2}{e^{2x}} + e \right) \cdot e^{2x}} = \frac{2e^2}{e^2 + e \cdot e^{2x}}$ Factor out $e$ from the denominator: $f(1-x) = \frac{2e^2}{e(e + e^{2x})} = \frac{2e}{e + e^{2x}}$ Now, let's compute $f(x) + f(1-x)$: $f(x) + f(1-x) = \frac{2e^{2x}}{e^{2x} + e} + \frac{2e}{e^{2x} + e}$ Since the denominators are the same, we can combine the numerators: $f(x) + f(1-x) = \frac{2e^{2x} + 2e}{e^{2x} + e} = \frac{2(e^{2x} + e)}{e^{2x} + e}$ Since $e^{2x} + e \neq 0$ for all real $x$, we can cancel the common factor: $f(x) + f(1-x) = 2$ This is a key result: the sum of the function evaluated at $x$ and $1-x$ is always 2.

Step 3: Apply the Functional Property to the Sum The sum is $S = f\left( \frac{1}{100} \right) + f\left( \frac{2}{100} \right) + \dots + f\left( \frac{99}{100} \right)$. We can pair the terms using the property $f(x) + f(1-x) = 2$. Pair the first term with the last: $f\left( \frac{1}{100} \right) + f\left( \frac{99}{100} \right) = f\left( \frac{1}{100} \right) + f\left( 1 - \frac{1}{100} \right) = 2$ Pair the second term with the second-to-last: $f\left( \frac{2}{100} \right) + f\left( \frac{98}{100} \right) = f\left( \frac{2}{100} \right) + f\left( 1 - \frac{2}{100} \right) = 2$ This pairing continues. The general pair is $f\left( \frac{k}{100} \right) + f\left( \frac{100-k}{100} \right) = 2$.

Step 4: Count the Number of Pairs and Identify the Middle Term The sum has 99 terms, from $k=1$ to $k=99$. The pairs are formed for $k=1, 2, \dots, 49$. For example, $k=1$ pairs with $k=99$, $k=2$ pairs with $k=98$, and $k=49$ pairs with $k=51$. Thus, there are 49 such pairs, and each pair sums to 2. The sum of these pairs is $49 \times 2 = 98$. Since there are 99 terms in total, and we have formed 49 pairs (which account for $49 \times 2 = 98$ terms), there is one term left in the middle. This middle term occurs when the argument is $1/2$. The middle term is $f\left( \frac{50}{100} \right) = f\left( \frac{1}{2} \right)$.

Step 5: Calculate the Value of the Middle Term We can find the value of the middle term $f(1/2)$ using the property $f(x) + f(1-x) = 2$. Substitute $x = 1/2$: $f\left( \frac{1}{2} \right) + f\left( 1 - \frac{1}{2} \right) = 2$ $f\left( \frac{1}{2} \right) + f\left( \frac{1}{2} \right) = 2$ $2f\left( \frac{1}{2} \right) = 2$ $f\left( \frac{1}{2} \right) = 1$ Alternatively, we can directly substitute $x=1/2$ into the function definition: $f\left(\frac{1}{2}\right) = \frac{2e^{2(1/2)}}{e^{2(1/2)} + e} = \frac{2e^1}{e^1 + e} = \frac{2e}{2e} = 1$

Step 6: Calculate the Total Sum The total sum $S$ is the sum of all the pairs plus the value of the middle term: $S = (\text{Sum of 49 pairs}) + (\text{Value of the middle term})$ $S = (49 \times 2) + 1$ $S = 98 + 1$ $S = 99$

Common Mistakes & Tips

• Algebraic Errors: Be meticulous with algebraic manipulations, especially when simplifying fractions and exponents. A small error can lead to an incorrect functional property.
• Missing the Middle Term: When the number of terms is odd, remember to account for the single middle term separately. If you only sum the pairs, your answer will be off by the value of the middle term.
• Recognizing Symmetry Early: The structure of the arguments ($k/100$ and $(100-k)/100$) is a strong hint to check for symmetry around $x=1/2$ or $f(x)+f(1-x)=C$.

Summary

The problem involves calculating a sum of function values where the arguments are in an arithmetic progression. The key to solving this efficiently is to recognize and derive the functional property $f(x) + f(1-x) = 2$. By pairing the terms in the sum such that their arguments add up to 1, we can group them into sets that sum to 2. Since there are 99 terms, we form 49 such pairs, and there is one middle term, $f(1/2)$, which evaluates to 1. The total sum is the sum of these pairs plus the middle term, resulting in $(49 \times 2) + 1 = 99$.

The final answer is $\boxed{99}$.