If f(x) = 2 ^ 2x 2 ^ 2x + 2 , x ∈ R , then f((1)/(2023)) +...

Given,

$f\left(x\right) = \dfrac{{2}^{2x}}{{2}^{2x} + 2}$,

so $f\left(1 - x\right) = \dfrac{{2}^{2 - 2x}}{{2}^{2 - 2x} + 2} = \dfrac{{2}^{2}}{{2}^{2} + {2}^{2x + 1}} = \dfrac{2}{2 + {2}^{2x}}$

Now $f\left(x\right) + f\left(1 - x\right) = \dfrac{{2}^{2x}}{{2}^{2x} + 2} + \dfrac{2}{2 + {2}^{2x}} = 1$

So by using the relation $f\left(x\right) + f\left(1 - x\right) = 1$ we will solve the given equation

$f\left(\dfrac{1}{2023}\right) + f\left(\dfrac{2}{2023}\right) + f\left(\dfrac{3}{2023}\right) . . . . . . . . . f\left(\dfrac{2022}{2023}\right)$

$= f\left(\dfrac{1}{2023}\right) + f\left(\dfrac{2022}{2023}\right) + f\left(\dfrac{2}{2023}\right) + f\left(\dfrac{2021}{2023}\right) + f\left(\dfrac{3}{2023}\right) + f\left(\dfrac{2020}{2023}\right) . . . . . . . . . f\left(\dfrac{1011}{2023}\right) + f\left(\dfrac{1012}{2023}\right)$

$= \mathop{\mathop{f\left(\dfrac{1}{2023}\right) + f\left(\dfrac{2022}{2023}\right)}_{⏟}}_{1} + \mathop{\mathop{f\left(\dfrac{2}{2023}\right) + f\left(\dfrac{2021}{2023}\right)}_{⏟}}_{1} + \mathop{\mathop{f\left(\dfrac{3}{2023}\right) + f\left(\dfrac{2020}{2023}\right)}_{⏟}}_{1} . . . . . . . . . + \mathop{\mathop{f\left(\dfrac{1011}{2023}\right) + f\left(\dfrac{1012}{2023}\right)}_{⏟}}_{1}$

$= 1 + 1 + 1 . . . . . . . . . 1011\text{times} = 1011$