\(A=1+3+3^2+3^3+...+3^{2014}\)
\(\Rightarrow3A=3+3^2+3^3+...+3^{2015}\)
\(\Rightarrow3A-A=\left(3+3^2+3^3+...+3^{2015}\right)-\left(1+3+3^2+...+3^{2014}\right)\)
\(\Rightarrow2A=3^{2015}-1\)
Ta có: \(2A+1=3^{2015}-1+1=3^{2015}⋮3\)
\(\Rightarrow2A+1⋮3\left(đpcm\right)\)