Ta có:
\(A=1+3+3^2+..+3^{2008}\)
\(\Rightarrow3A=\left(1+3+3^2+...+3^{2008}\right).3\)
\(=3+3^2+3^3...+3^{2009}\)
Vì \(2A=3A-A\)nên ta có:
\(2A=\left(3+3^2+3^3+...+3^{2009}\right)-\left(1+3+3^2+..+3^{2008}\right)\)
\(=3^{2009}-1\)
\(\Rightarrow2A-3^{2009}=3^{2009}-1-3^{2009}\)
\(=\left(3^{2009}-3^{2009}\right)-1\)
\(=0-1\)
\(=-1\)
Vậy \(2A-3^{2009}=-1\)