\(A=1+3+3^2+...+3^{2008}\\ 3A=3\left(1+3+3^2+...+3^{2008}\right)\\ 3A=3+3^2+3^3+...+3^{2009}\\ 3A-A=\left(3+3^2+3^3+...+3^{2009}\right)-\left(1+3+3^2+...+3^{2008}\right)\\ 2A=3^{2009}-1\\ 2A+1=3^{2009}-1+1\\ 2A+1=3^{2009}\\ \dfrac{2A+1}{3}=\dfrac{3^{2009}}{3}\\ \dfrac{2A+1}{3}=3^{2008}=3^{1004\cdot2}=\left(3^{1004}\right)^2\)
Vậy \(\dfrac{2A+1}{3}\) là số chính phương