Ta có:
\(A=1+3+3^2+.........+3^{2000}\)
\(\Rightarrow3.A=3+3^2+3^3+...........+3^{2001}\)
Khi đó: \(3.A-A=\left(3+3^2+3^3+......+3^{2001}\right)-\left(1+3+3^2+......+3^{2000}\right)\)
\(\Rightarrow2.A=3^{2001}-1\)
\(\Rightarrow n=2001\)
Vậy: n = 2001.
\(3A=3+3^2+...+3^{2001}\)
\(\Rightarrow3A-A=\left(3+3^2+...+3^{2001}\right)-\left(1+3+...+3^{2000}\right)\)
\(\Rightarrow2A=3^{2001}-1\)
\(\Rightarrow3^{2001}-1=3^n-1\)
\(\Rightarrow3^{2001}=3^n\)
\(\Rightarrow n=2001\)
Vậy n = 2001