Câu hỏi của Đỗ Nam Trâm

Question

S=$2+2^2+2^3+...+2^{\text{2010}}$ chia hết cho 7,31

Nguyễn Việt Lâm · Accepted Answer

$S=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{2008}\left(1+2+2^2\right)$$=2.7+2^4.7+...+2^{2008}.7$$=7\left(2+2^4+...+2^{2008}\right)⋮7$$S=2\left(1+2+2^2+2^3+2^4\right)+2^6\left(1+2+2^2+2^3+2^4\right)+...+2^{2006}\left(1+2+2^2+2^3+2^4\right)$$=2.31+2^6.31+...+2^{2006}.31$$=31\left(2+2^6+...+2^{2006}\right)⋮31$Câu trả lời: $S=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{2008}\left(1+2+2^2\right)$$=2.7+2^4.7+...+2^{2008}.7$$=7\left(2+2^4+...+2^{2008}\right)⋮7$$S=2\left(1+2+2^2+2^3+2^4\right)+2^6\left(1+2+2^2+2^3+2^4\right)+...+2^{2006}\left(1+2+2^2+2^3+2^4\right)$$=2.31+2^6.31+...+2^{2006}.31$$=31\left(2+2^6+...+2^{2006}\right)⋮31$

Nguyễn Lê Phước Thịnh · Answer

Ta có: $S=2+2^2+2^3+...+2^{2010}$$=\left(2+2^2+2^3+2^4+2^5+2^6\right)+...+\left(2^{2005}+2^{2006}+2^{2007}+2^{2008}+2^{2009}+2^{2010}\right)$$=126\cdot\left(1+...+2^{2005}\right)⋮7$Ta có: $S=2+2^2+2^3+...+2^{2010}$$=\left(2+2^2+2^3+2^4+2^5\right)+...+\left(2^{2006}+2^{2007}+2^{2008}+2^{2009}+2^{2010}\right)$$=31\cdot2\cdot\left(1+...+2^{2006}\right)⋮31$Câu trả lời: Ta có: $S=2+2^2+2^3+...+2^{2010}$$=\left(2+2^2+2^3+2^4+2^5+2^6\right)+...+\left(2^{2005}+2^{2006}+2^{2007}+2^{2008}+2^{2009}+2^{2010}\right)$$=126\cdot\left(1+...+2^{2005}\right)⋮7$Ta có: $S=2+2^2+2^3+...+2^{2010}$$=\left(2+2^2+2^3+2^4+2^5\right)+...+\left(2^{2006}+2^{2007}+2^{2008}+2^{2009}+2^{2010}\right)$$=31\cdot2\cdot\left(1+...+2^{2006}\right)⋮31$

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