Key Concepts and Formulas

• Functional Property: Investigating properties like $f(x) + f(a-x) = C$ is crucial for simplifying sums of function values, especially when the arguments form an arithmetic progression symmetric around $a/2$.
• Exponent Rules: The rule $a^{m-n} = \frac{a^m}{a^n}$ is frequently used to manipulate terms involving exponents.
• Arithmetic Series Sum: The sum of an arithmetic series $1 + 2 + \dots + n = \frac{n(n+1)}{2}$.

Step-by-Step Solution

Step 1: Analyze the function and identify potential symmetry. The given function is $f(x) = \frac{2^{2x}}{2^{2x} + 2}$. We can rewrite $2^{2x}$ as $(2^2)^x = 4^x$. So, $f(x) = \frac{4^x}{4^x + 2}$. The sum involves terms $f\left(\frac{1}{2023}\right), f\left(\frac{2}{2023}\right), \dots, f\left(\frac{2022}{2023}\right)$. Notice that the arguments are of the form $\frac{k}{2023}$ where $k$ ranges from 1 to 2022. The sum of the first and last argument is $\frac{1}{2023} + \frac{2022}{2023} = \frac{2023}{2023} = 1$. This suggests investigating the property $f(x) + f(1-x)$.

Step 2: Derive the functional property $f(x) + f(1-x)$. We have $f(x) = \frac{4^x}{4^x + 2}$. To find $f(1-x)$, substitute $(1-x)$ for $x$: $f(1-x) = \frac{4^{1-x}}{4^{1-x} + 2}$ Using the exponent rule $4^{1-x} = \frac{4^1}{4^x} = \frac{4}{4^x}$, we get: $f(1-x) = \frac{\frac{4}{4^x}}{\frac{4}{4^x} + 2}$ To simplify the denominator of $f(1-x)$: $\frac{4}{4^x} + 2 = \frac{4}{4^x} + \frac{2 \cdot 4^x}{4^x} = \frac{4 + 2 \cdot 4^x}{4^x}$ So, $f(1-x)$ becomes: $f(1-x) = \frac{\frac{4}{4^x}}{\frac{4 + 2 \cdot 4^x}{4^x}} = \frac{4}{4^x} \cdot \frac{4^x}{4 + 2 \cdot 4^x} = \frac{4}{4 + 2 \cdot 4^x}$ We can simplify this by dividing the numerator and denominator by 2: $f(1-x) = \frac{2}{2 + 4^x}$ Now, let's find $f(x) + f(1-x)$: $f(x) + f(1-x) = \frac{4^x}{4^x + 2} + \frac{2}{4^x + 2}$ Since the denominators are the same, we can add the numerators: $f(x) + f(1-x) = \frac{4^x + 2}{4^x + 2} = 1$ So, we have established the property $f(x) + f(1-x) = 1$.

Step 3: Apply the functional property to the given sum. The sum is $S = f\left(\frac{1}{2023}\right) + f\left(\frac{2}{2023}\right) + \dots + f\left(\frac{2022}{2023}\right)$. Let's pair terms from the beginning and the end of the sum. The first term is $f\left(\frac{1}{2023}\right)$. Its complementary term, based on the property $f(x) + f(1-x) = 1$, would be $f\left(1 - \frac{1}{2023}\right) = f\left(\frac{2022}{2023}\right)$. The sum of this pair is $f\left(\frac{1}{2023}\right) + f\left(\frac{2022}{2023}\right) = 1$.

The second term is $f\left(\frac{2}{2023}\right)$. Its complementary term is $f\left(1 - \frac{2}{2023}\right) = f\left(\frac{2021}{2023}\right)$. The sum of this pair is $f\left(\frac{2}{2023}\right) + f\left(\frac{2021}{2023}\right) = 1$.

This pattern continues. The general pair is $f\left(\frac{k}{2023}\right) + f\left(\frac{2023-k}{2023}\right) = 1$.

Step 4: Count the number of pairs and the middle term (if any). The sum has $2022$ terms. The terms are $f\left(\frac{1}{2023}\right), f\left(\frac{2}{2023}\right), \dots, f\left(\frac{2022}{2023}\right)$. We are pairing $\frac{k}{2023}$ with $\frac{2023-k}{2023}$. The pairing stops when $k = 2023-k$, which means $2k = 2023$. This gives $k = \frac{2023}{2}$, which is not an integer. Therefore, there is no middle term that pairs with itself. The number of terms is $2022$. We are forming pairs of the form $f(x) + f(1-x)$. The number of such pairs is $\frac{2022}{2} = 1011$. Each pair sums to 1.

Step 5: Calculate the total sum. Since there are 1011 pairs, and each pair sums to 1, the total sum is: $S = \underbrace{1 + 1 + \dots + 1}_{1011 \text{ times}}$ $S = 1011 \times 1 = 1011$

Common Mistakes & Tips

• Incorrect Pairing: Ensure that the arguments of the paired terms sum to the constant (in this case, 1). For example, $\frac{k}{n}$ pairs with $\frac{n-k}{n}$.
• Missing Middle Term: If the total number of terms is odd, there might be a middle term that does not form a pair. In this problem, the number of terms (2022) is even, so all terms are paired.
• Algebraic Errors: Be meticulous with algebraic manipulations, especially when dealing with exponents and fractions. Double-checking the derivation of $f(x) + f(1-x)$ is crucial.

Summary

The problem involves calculating a sum of function values. By analyzing the function $f(x) = \frac{2^{2x}}{2^{2x} + 2}$ and the arguments of the terms in the sum, we identified a key functional property: $f(x) + f(1-x) = 1$. This property allows us to pair terms in the sum. The sum consists of 2022 terms. We can form $\frac{2022}{2} = 1011$ pairs, where each pair $f\left(\frac{k}{2023}\right) + f\left(\frac{2023-k}{2023}\right)$ sums to 1. Therefore, the total sum is $1011 \times 1 = 1011$.

The final answer is \boxed{1011}. which corresponds to option (D).