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

$$ \begin{aligned} & f(x)=\frac{4^x}{4^x+2} \\\\ & f(x)+f(1-x)=\frac{4^x}{4^x+2}+\frac{4^{1-x}}{4^{1-x}+2} \\\\ & =\frac{4^x}{4^x+2}+\frac{4}{4+2\left(4^x\right)} \\\\ & =\frac{4^x}{4^{\mathrm{x}}+2}+\frac{2}{2+4^{\mathrm{x}}} \\\\ & =1 \\\\ & \Rightarrow \mathrm{f}(\mathrm{x})+\mathrm{f}(1-\mathrm{x})=1 \\\\ & \text { Now } \mathrm{f}\left(\frac{1}{2023}\right)+\mathrm{f}\left(\frac{2}{2023}\right)+\mathrm{f}\left(\frac{3}{2023}\right)+\ldots \ldots .+ \\\\ & \ldots \ldots \ldots . .+\mathrm{f}\left(1-\frac{3}{2023}\right)+\mathrm{f}\left(1-\frac{2}{2023}\right)+\mathrm{f}\left(1-\frac{1}{2023}\right) \end{aligned} $$<br/><br/> Now sum of terms equidistant from beginning and end is 1<br/><br/> $$ \begin{aligned} & \text { Sum }=1+1+1+\ldots \ldots \ldots+1 \text { (1011 times }) \\\\ & =1011 \end{aligned} $$