Xet \(n=3k\)
\(\Rightarrow3^{6k}+3^{3k}+1\equiv3\left(mod13\right)\)
Xet \(n=3k+1\)
\(\Rightarrow3^{6k+2}+3^{3k+1}+1\equiv9+3+1\equiv0\left(mod13\right)\)
Xet \(n=3k+2\)
\(\Rightarrow3^{6k+3+1}+3^{3k+2}+1\equiv3+9+1\equiv0\left(mod13\right)\)
Vậy vơi mọi n tự nhiên và n không chia hêt cho 3 thì
\(3^{2n}+3^n+1⋮13\)