Câu hỏi của Bảo Nhật Trương

Question

Chứng minh:D=1+2+2 mũ 2 +...+2 mũ 2008 chia 7 dư 3

Nguyễn Ngọc Anh Minh · Accepted Answer

$D=3+2^2...............2^{2008}$

$B=\left(2^2+2^3+2^4\right)+\left(2^5+2^6+2^7\right)+...+\left(2^{2006}+2^{2007}+2^{2008}\right)=$

$=2^2\left(1+2+2^2\right)+2^5\left(1+2+2^2\right)+2^{2006}\left(1+2+2^2\right)=$

$=7\left(2^2+2^5+2^8+...+2^{2003}+2^{2006}\right)⋮7$

$D=3+B$ mà $B⋮7$ => D chia 7 dư 3Câu trả lời: $D=3+2^2...............2^{2008}$

$B=\left(2^2+2^3+2^4\right)+\left(2^5+2^6+2^7\right)+...+\left(2^{2006}+2^{2007}+2^{2008}\right)=$

$=2^2\left(1+2+2^2\right)+2^5\left(1+2+2^2\right)+2^{2006}\left(1+2+2^2\right)=$

$=7\left(2^2+2^5+2^8+...+2^{2003}+2^{2006}\right)⋮7$

$D=3+B$ mà $B⋮7$ => D chia 7 dư 3