ta có : \(A=1+3+3^2+...+3^{99}\)
\(\Rightarrow3A=3\left(1+3+3^2+...+3^{99}\right)\)
\(\Leftrightarrow3A=3+3^2+3^3+...+3^{100}\)
\(\Rightarrow3A-A=2A=\left(3+3^2+3^3+...+3^{100}\right)-\left(1+3+3^2+...+3^{99}\right)\)
\(\Leftrightarrow2A=3^{100}-1\)
\(\Rightarrow2A+1=3^{100}-1+1=3^{100}=\left(3^{25}\right)^4\)
vậy \(2A+1=\left(3^{25}\right)^4\)