Ta có:
\(A=3+3^2+3^3+...+3^{10}\)
\(\Rightarrow3A=3^2+3^3+3^4+...+3^{11}\)
\(\Rightarrow3A-A=\left(3^2+3^3+3^4+...+3^{11}\right)-\left(3+3^2+3^3+...+3^{10}\right)\)
\(\Rightarrow2A=3^{11}-3\)
\(\Rightarrow2A+3=3^{11}-3+3\)
\(\Rightarrow2A+3=3^{11}\)
Vậy \(2A+3=3^{11}\)