If [f(x) = 2^2x 2^2x + 2,x R], then [f( 1 2023 ) + f( 2 2023 ), + ,..., + ,f( 2022 2023 )]…

Question

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

Accepted Answer

$$
\begin{aligned}
& f(x)=\frac{4^x}{4^x+2} \\
& f(x)+f(1-x)=\frac{4^x}{4^x+2}+\frac{4^{1-x}}{4^{1-x}+2} \\
& =\frac{4^x}{4^x+2}+\frac{4}{4+2\left(4^x\right)} \\
& =\frac{4^x}{4^{\mathrm{x}}+2}+\frac{2}{2+4^{\mathrm{x}}} \\
& =1 \\
& \Rightarrow \mathrm{f}(\mathrm{x})+\mathrm{f}(1-\mathrm{x})=1 \\
& \text { Now } \mathrm{f}\left(\frac{1}{2023}\right)+\mathrm{f}\left(\frac{2}{2023}\right)+\mathrm{f}\left(\frac{3}{2023}\right)+\ldots \ldots .+ \\
& \ldots \ldots \ldots . .+\mathrm{f}\left(1-\frac{3}{2023}\right)+\mathrm{f}\left(1-\frac{2}{2023}\right)+\mathrm{f}\left(1-\frac{1}{2023}\right)
\end{aligned}
$$

Now sum of terms equidistant from beginning and end is 1

$$
\begin{aligned}
& \text { Sum }=1+1+1+\ldots \ldots \ldots+1 \text { (1011 times }) \\
& =1011
\end{aligned}
$$

Solutions · 1