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$f(x)+f(1-x)=\frac{100^x}{100^x + 10}+\frac{100^{1-x}}{100^{1-x} + 10}=\frac{100^x}{100^x + 10}+\frac{100}{10 \cdot 100^{x} + 100}=1$ Thus, everything cancels to get \[ f\left( \frac{1}{2024} \right) - f\left( \frac{2}{2024} \right) + f\left( \frac{3}{2024} \right) - f\left( \frac{4}{2024} \right) + \ldots - f\left( \frac{2022}{2024} \right) + f\left( \frac{2023}{2024} \right)=1-f\left( \frac{1012}{2024} \right)=\frac{1}{2} \]

1012 terms: A = f (1/2024) + f (3/2024) + ... + f (2021/2024) + f (2023/2024) 1011 terms: B = f (2/2024) + f (4/2024) +. .. + f (2022/2024) f (x) + f (1-x) = 1 A: 506 pairs of 1 B: 505 pairs of 1 and term 1012 2 f(1012/2024) =1, f(1012/2024) = 1/2 A - B = 506 - 505 - 1/2 = 1/2 Please check this out.

Yes that is correct, I have fixed my post above.

Thanks for your attention.