Question

If $f(x)=\frac{2^{2x}}{2^{2x}+2}, x\in R$, then $f(\frac{1}{2023})+f(\frac{2}{2023})+...+f(\frac{2022}{2023})$ is equal to:

• A. 2011
• B. 2010
• C. 1010
• D. 101

Answer 1

Answer: (C)

1010

Answer 2

Let $f(x)=\frac{2^{2x}}{2^{2x}+2}$. We observe that $f(x)+f(1-x) = \frac{2^{2x}}{2^{2x}+2} + \frac{2^{2(1-x)}}{2^{2(1-x)}+2}$. $f(1-x) = \frac{2^{2-2x}}{2^{2-2x}+2} = \frac{4 \cdot 2^{-2x}}{4 \cdot 2^{-2x}+2}$. Multiplying numerator and denominator by $2^{2x}$: $f(1-x) = \frac{4}{4+2 \cdot 2^{2x}} = \frac{2}{2+2^{2x}}$. So, $f(x)+f(1-x) = \frac{2^{2x}}{2^{2x}+2} + \frac{2}{2^{2x}+2} = \frac{2^{2x}+2}{2^{2x}+2} = 1$.

The sum is $S = f(\frac{1}{2023})+f(\frac{2}{2023})+...+f(\frac{2022}{2023})$. Let $n=2023$. The sum is $\sum_{k=1}^{n-1} f(\frac{k}{n})$. We can pair the terms: $f(\frac{k}{n}) + f(\frac{n-k}{n}) = f(\frac{k}{n}) + f(1-\frac{k}{n}) = 1$. There are $n-1 = 2022$ terms. The number of pairs is $\frac{n-1}{2} = \frac{2022}{2} = 1011$. Each pair sums to 1. Therefore, the total sum is $1011 \times 1 = 1011$.

However, 1011 is not an option. This suggests a potential typo in the question or options. If we assume the question intended the sum to be up to $f(\frac{2020}{2021})$, then $n=2021$. The sum would be $\sum_{k=1}^{2020} f(\frac{k}{2021})$. Number of terms = 2020. Number of pairs = $\frac{2020}{2} = 1010$. Each pair sums to 1. Total sum = $1010 \times 1 = 1010$. This matches option (3). Assuming this intended question, the answer is 1010.