\(Q=2+2^2+2^3+...+2^{99}\)
\(=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{98}+2^{99}\right)\)
\(=6+2^2.\left(2+2^2\right)+...+2^{97}.\left(2+2^2\right)\)
\(=6+2^2.6+...+2^{97}.6\)
\(=6.\left(1+2^2+...+2^{97}\right)\)
Vì \(6⋮3\) nên \(6.\left(1+2^2+...+2^{97}\right)⋮3\)
Vậy \(Q⋮3\)
\(#WendyDang\)