Ta có: \(A=1-3+3^2-3^3+\cdots+3^{2020}-3^{2021}\)
=>3A=\(3-3^2+3^3-3^4+\cdots+3^{2021}-3^{2022}\)
=>3A+A=\(3-3^2+3^3-3^4+\cdots+3^{2021}-3^{2022}+1-3+3^2-\cdots+3^{2020}-3^{2021}\)
=>4A=\(-3^{2022}+1\)
\(4A+3^{n}=1\)
=>\(-3^{2022}+1+3^{n}=1\)
=>\(3^{n}=3^{2022}\)
=>n=2022
