Câu hỏi của Trần Mạnh Tiến

Question

CHO M =$2^0+2^2+2^4+2^6+.+2^{2018}$TÌM SỐ DƯ KHI M CHIA CHO 7HELP ME !!!

Phan Minh Thiện · Accepted Answer

$M=2^0+2^2+2^4+2^6+2^8+...+2^{2018}$$M=2^0+2^2+\left(2^4+2^6+2^8\right)+...+\left(2^{2014}+2^{2016}+2^{2018}\right)$$M=1+4+2^4.\left(1+2^2+2^4\right)+...+2^{2014}.\left(1+2^2+2^4\right)$$M=5+2^4.21+2^{10}.21+...+2^{2014}.21$$M=5+21.\left(2^4+2^{10}+...+2^{2014}\right)$vì  $21.\left(2^4+2^{10}+...+2^{2014}\right)⋮7$nên $M=5+21.\left(2^4+2^{10}+...+2^{2014}\right)$chia 7 dư 5Câu trả lời: $M=2^0+2^2+2^4+2^6+2^8+...+2^{2018}$$M=2^0+2^2+\left(2^4+2^6+2^8\right)+...+\left(2^{2014}+2^{2016}+2^{2018}\right)$$M=1+4+2^4.\left(1+2^2+2^4\right)+...+2^{2014}.\left(1+2^2+2^4\right)$$M=5+2^4.21+2^{10}.21+...+2^{2014}.21$$M=5+21.\left(2^4+2^{10}+...+2^{2014}\right)$vì  $21.\left(2^4+2^{10}+...+2^{2014}\right)⋮7$nên $M=5+21.\left(2^4+2^{10}+...+2^{2014}\right)$chia 7 dư 5

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