Câu hỏi của Vũ Đức Duy Anh

Question

cho A =2+2mũ 2 +2 mũ 3 +.....+ 2 mũ 60 chứng minh A chia hết cho 3 ,5,7

Đoàn Đức Hà · Accepted Answer

$A=2+2^2+2^3+...+2^{60}$$=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{59}+2^{60}\right)$$=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{59}\left(1+2\right)$$=3\left(2+2^3+...+2^{59}\right)$chia hết cho $3$.$A=2+2^2+2^3+...+2^{60}$$=\left(2+2^2+2^3+2^4\right)+...+\left(2^{57}+2^{58}+2^{59}+2^{60}\right)$$=2\left(1+2+2^2+2^3\right)+...+2^{57}\left(1+2+2^2+2^3\right)$$=15\left(2+...+2^{57}\right)⋮5$$A=2+2^2+2^3+...+2^{60}$$=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{58}+2^{59}+2^{60}\right)$$=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{58}\left(1+2+2^2\right)$$=7\left(2+2^4+...+2^{58}\right)$chia hết cho $7$.Câu trả lời: $A=2+2^2+2^3+...+2^{60}$$=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{59}+2^{60}\right)$$=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{59}\left(1+2\right)$$=3\left(2+2^3+...+2^{59}\right)$chia hết cho $3$.$A=2+2^2+2^3+...+2^{60}$$=\left(2+2^2+2^3+2^4\right)+...+\left(2^{57}+2^{58}+2^{59}+2^{60}\right)$$=2\left(1+2+2^2+2^3\right)+...+2^{57}\left(1+2+2^2+2^3\right)$$=15\left(2+...+2^{57}\right)⋮5$$A=2+2^2+2^3+...+2^{60}$$=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{58}+2^{59}+2^{60}\right)$$=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{58}\left(1+2+2^2\right)$$=7\left(2+2^4+...+2^{58}\right)$chia hết cho $7$.

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