\(1+2+2^2+...+2^{2009}+2^{2010}\)
\(1+\left(2+2^2+2^3\right)+...+\left(2^{2008}+2^{2009}+2^{2010}\right)\)
=\(1+2\left(1+2+2^2\right)+...+2^{2008}\left(1+2+2^2\right)\)
=\(1+\left(2+2^4+...+2^{2008}\right)\left(1+2+2^2\right)\)
=\(1+\left(2+2^4+...+2^{2008}\right)7\)
=>\(1+2+2^2+...+2^{2009}+2^{2010}\) chia cho 7 dư 1