đề sai em ơi số cuối phải là 2^2009
\(A=1+2+2^2+2^3+...+2^{2008}+2^{2009}\)
\(A=\left(1+2+2^2\right)+\left(2^3+2^4+2^5\right)+...+\left(2^{2007}+2^{2008}+2^{2009}\right)\)
\(A=7+2^3\left(1+2+2^2\right)+...+2^{2007}\left(1+2+2^2\right)\)
\(A=7\cdot1+7\cdot2^3+...+7\cdot2^{2007}\)
\(A=7\left(1+2^3+...+2^{2007}\right)⋮7\)
=> A chia 7 dư 0
\(A=\)nhưu trên
=>\(A=1+2+\left(2^2+2^3+2^4\right)+...+\left(2^{2006}+2^{2007}+2^{2008}\right)\)(có 669 nhóm and thừa 2 số)
=>\(A=3+2^2\left(1+2+2^2\right)+...+2^{2006}\left(1+2+2^2\right)\)
=>\(A=3+\left(1+2+2^2\right)\left(2^5+...+2^{2006}\right)\)
=>\(A=3+7\left(2^5+...+2^{2006}\right)\)
=>\(A\)chia cho 7 dư 3
A chia cho 7 dư 3 nhé
hoktot