Đáp án:
$x=504$
Giải thích các bước giải:
$C=4+2^2+2^3\ +\,.\!.\!.+\ 2^{2021}\\\Rightarrow 2C=2.(2^2+2^2+2^3\ +\,.\!.\!.+\ 2^{2021})\\\Rightarrow 2C=2^3+2^3+2^4\ +\,.\!.\!.+\ 2^{2022}\\\Rightarrow 2C-C=(2^3+2^3+2^4\ +\,.\!.\!.+\ 2^{2022})-(2^2+2^2+2^3\ +\,.\!.\!.+\ 2^{2021})\\\Rightarrow C=2^{2022}+2^3-(2^2+2^2)\\\Rightarrow C=2^{2022}+2^3-2^3\\\Rightarrow C=2^{2022}\\\Rightarrow 2^2. C=2^2. 2^{2022}\\\Rightarrow 2^2. C=2^{2024}\\\Rightarrow 16^{x+2}=2^{2024}\\\Rightarrow \left(2^4\right)^{x+2}=2^{2024}\\\Rightarrow 2^{4x+8}=2^{2024}\\\Rightarrow 4x+8=2024\\\Rightarrow 4x=2016\\\Rightarrow x=504$
