india.tour.edu.vn

If f(x)=(2^(2x))/(2^(2x)+2),x in z, then f((1)/(2023))+...+f((2022)/(2

05:42 09/07/2025

To solve the problem, we need to evaluate the sum f(12023)+f(22023)+…+f(20222023) where f(x)=22x22x+2.Step 1: Simplify the function \( f(x) \)We start with the function: \( f(x) = \frac{2^{2x}}{2^{2x} + 2} \)We can rewrite \( 2^{2x} \) as \( 4^x \): \( f(x) = \frac{4^x}{4^x + 2} \)Step 2: Find \( f(1-x) \)Next, we will find \( f(1-x) \): \( f(1-x) = \frac{4^{1-x}}{4^{1-x} + 2} = \frac{\frac{4}{4^x}}{\frac{4}{4^x} + 2} = \frac{4}{4 + 2 \cdot 4^x} \)Step 3: Compute \( f(x) + f(1-x) \)Now we will compute \( f(x) + f(1-x) \): \( f(x) + f(1-x) = \frac{4^x}{4^x + 2} + \frac{4}{4 + 2 \cdot 4^x} \)To combine these fractions, we find a common denominator: \( = \frac{4^x(4 + 2 \cdot 4^x) + 4(4^x + 2)}{(4^x + 2)(4 + 2 \cdot 4^x)} \)Step 4: Simplify the numeratorExpanding the numerator: \( = \frac{4^{x+1} + 2 \cdot 4^{2x} + 4^{x+1} + 8}{(4^x + 2)(4 + 2 \cdot 4^x)} \) \( = \frac{2 \cdot 4^{x+1} + 2 \cdot 4^{2x} + 8}{(4^x + 2)(4 + 2 \cdot 4^x)} \) Factoring out 2: \( = \frac{2(4^{x+1} + 4^{2x} + 4)}{(4^x + 2)(4 + 2 \cdot 4^x)} \)Step 5: Evaluate \( f(x) + f(1-x) \)Notice that \( f(x) + f(1-x) = 1 \): \( f(x) + f(1-x) = 1 \)Step 6: Sum \( f\left(\frac{k}{2023}\right) \)Now, we can compute the sum: \( f\left(\frac{1}{2023}\right) + f\left(\frac{2}{2023}\right) + \ldots + f\left(\frac{2022}{2023}\right) \)We can pair the terms: \( f\left(\frac{k}{2023}\right) + f\left(\frac{2023-k}{2023}\right) = 1 \)Since there are 2022 terms, we can pair them up: \( \text{Number of pairs} = \frac{2022}{2} = 1011 \)Step 7: Final ResultThus, the total sum is: \( 1011 \cdot 1 = 1011 \)ConclusionThe final answer is: \( \boxed{1011} \)