Thank you for reporting, we will resolve it shortly
Q.
If $f(x)=\frac{2^{2 x}}{2^{2 x}+2}, x \in R$, then $f\left(\frac{1}{2023}\right)+f\left(\frac{2}{2023}\right)+\ldots+f\left(\frac{2022}{2023}\right)$ is equal to
$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^x+2}+\frac{2}{2+4^x}$
$ =1 $
$ \Rightarrow f(x)+f(1-x)=1$
Now $ f \left(\frac{1}{2023}\right)+ f \left(\frac{2}{2023}\right)+ f \left(\frac{3}{2023}\right)+\ldots \ldots+ $
$ \ldots \ldots \ldots+ f \left(1-\frac{3}{2023}\right)+ f \left(1-\frac{2}{2023}\right)+ f \left(1-\frac{1}{2023}\right)$
Now sum of terms equidistant from beginning and end is $1$
Sum $=1+1+1+\ldots \ldots \ldots .+1(1011$ times $)$
$=1011$