\(T=3+3^2+3^3+...+3^{99}\)
\(3T=3\left(3+3^2+...+3^{99}\right)\)
\(3T=3^2+3^3+...+3^{100}\)
\(3T-T=\left(3^2+3^3+...+3^{100}\right)-\left(3+3^2+...+3^{99}\right)\)
\(2T=3^{100}-3\)
Thay vào ta có:\(3^{100}-3+3=3^{2n}\)
\(\Rightarrow3^{100}=3^{2n}\)
\(\Rightarrow100=2n\)
\(\Rightarrow n=50\)