\(A=2+2^2+2^3+2^4+...+2^{2008}\)
\(A=\left(2+2^2+2^3+2^4\right)+...+\left(2^{2005}+2^{2006}+2^{2007}+2^{2008}\right)\)
\(A=30+...+30\cdot2^{2004}\)
\(A=30\left(1+...+2^{2004}\right)⋮30\)
A = 2 + 22 + 23 + 24 + ... + 22007 + 22008
A = (2 + 22 + 23 + 24) + ... + (22005 + 22006 + 22007 + 22008)
A = 30 + ... + 22004.(2 + 22 + 23 + 24)
A = 30 + ... + 22004.30
A = 30.( 1 + ... + 22004) \(⋮\)30
ta có: A = 2 + 22 + 23 + 24 +...+ 22007 + 22008 ( có 2008 số hạng; 2008 : 4 = 502)
A = (2+22 + 23 +24)+...+ (22005+22006+22007+22008)
A = 30 + ...+22004.(2+22 +23 +24)
A = 30.(1+...+2004) chia hết cho 30