Ta có
\(B=\left(1+2^3\right)+2^2\left(1+2^3\right)+......+2^8\left(1+2^3\right)=9\left(1+2^2+....+2^8\right)\) Chia hết cho 9
(đpcm)
\( B=1+2+2^2+2^3+...+2^{11}\)
\(\Leftrightarrow B=\left(1+2^3\right)+2^2\left(1+2^3\right)+2^3\left(1+2^3\right)+...+2^8\left(1+2^3\right)\)
\(\Leftrightarrow B=9+2^2.9+2^3.9+...+2^8.9\)
\(\Leftrightarrow B=9\left(1+2^2+2^3+...+2^8\right)\)
Vậy \(B⋮9\left(ĐPCM\right)\)