\(B=2+2^2+2^3+2^4+...+2^9+2^{10}\)
\(2B=2^2+2^3+2^4+...+2^{10}+2^{11}\). Do 2B - B = B nên
\(B=\left(2^2+2^3+2^4+...+2^{10}+2^{11}\right)-\left(2+2^2+2^3+2^4+...+2^9+2^{10}\right)\)
\(=2^{11}-2⋮3^{\left(đpcm\right)}\)
\(B=2+2^2+2^3+2^4+...+2^9+2^{10}\)
\(B=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^9+2^{10}\right)\)
\(B=2\left(1+2\right)+2^3\left(1+2\right)+...+2^9\left(1+2\right)\)
\(B=2.3+2^3.3+...+2^9.3\)
\(B=3\left(2+2^3+...+2^9\right)⋮3\) ( đpcm )
Vậy \(B⋮3\)