Ta có: \(A=1+2+2^2+....+2^{2005}\)
\(=1+\left(2+2^2+2^3\right)+.....+\left(2^{2003}+2^{2004}+2^{2005}\right)\)
\(=1+2.\left(1+2+2^2\right)+.....+2^{2003}.\left(1+2+2^2\right)\)
\(=1+2.7+.....+2^{2003}.7\)
\(1+7.\left(2+.....+2^{2003}\right)\)
Vì \(7.\left(2+....+2^{2003}\right)\) chia hết cho 7
=> A chia cho 7 dư 1