Đặt A = 2 + 22 + 23 + ... + 22004
2A = 22 + 23 + 24 + ... + 22005
2A - A = (22 + 23 + 24 + ... + 22005) - (2 + 22 + 23 + ... + 22004)
A = 22005 - 2
Ta có: \(2^6\equiv1\left(mod21\right)\)
=> \(2^{2004}\equiv1\left(mod21\right)\)
=> 22004 - 1 chia hết cho 21
=> 2.(22004 - 1) chia hết cho 42
=> 22005 - 2 chia hết cho 42
=> A chia hết cho 42 (đpcm)
\(\left(2+2^2+2^3+2^4+2^5+2^6\right)+2^5\left(2+2^2+2^3+2^4+2^5+2^6\right)+...+2^{334}\left(2+2^2+2^3+2^4+2^5+2^6\right)\)
=\(126+2^5.126+...+2^{334}.126=126\left(1+2^5+2^{11}+...+2^{334}\right)\) chia hết cho 126 hay 42