\(A=3+3^2+3^3+...+3^{100}\)
\(\Rightarrow3A=3^2+3^3+3^4+...+3^{100}+3^{101}\)
\(\Rightarrow3A-A=\left(3^2+3^3+3^4+...+3^{100}+3^{101}\right)-\left(3+3^2+3^3+...+3^{100}\right)\)\(\Rightarrow2A=3^{101}-3\)
\(\Rightarrow2A+3=3^{101}-3+3=3^{101}\)
Vậy \(n=101\)