\(3+3^2+.....+3^{99}\)
\(=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{97}+3^{98}+3^{99}\right)\)
\(=39+3^3\left(3+3^2+3^3\right)+........+3^{96}\left(3+3^2+3^3\right)\)
\(=39+3^3\cdot39+...+3^{96}\cdot39\)
\(=39\left(1+3^3+....+3^{96}\right)\)
Vì \(39⋮13\Rightarrow39\in B\left(13\right)\)
Ý là bội của 13 đó bạn