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Chứng minh: 1+2+2^2+2^3+....+2^12+2^13+2^14 chia hết cho 31

 

[Blue Plus]

Chứng minh: 1+2+2^2+2^3+....+2^12+2^13+2^14 chia hết cho 31

[tex]\dpi{100} 1+2+2^2+2^3+...+2^{14}\\=(1+2+2^2+2^3+2^4)+(2^5+2^6+2^7+2^8+2^9)+(2^{10}+2^{11}+2^{12}+2^{13}+2^{14})\\=1(1+2+2^2+2^3+2^4)+2^5(1+2+2^2+2^3+2^4)+2^{10}(1+2+2^2+2^3+2^4)\\=(1+2+2^2+2^3+2^4)(1+2^5+2^{10})\\=31.(1+2^5+2^{10})\vdots 31\\ VAY1+2+2^2+2^3+...+2^{14}\vdots 31[/tex]

 

[orangery]

$A= 1+2+2^2+2^3+....+2^{12}+2^{13}+2^{14} \\
A= (1+2+2^2+2^3+2^4) + (2^5+2^6+2^7+2^8+2^9)+(2^{10}+2^{11}+2^{12}+2^{13}+2^{14})\\
A= (1+2+2^2+2^3+2^4) + 2^5.(1+2+2^2+2^3+2^4) + 2^{10}.(1+2+2^2+2^3+2^4) \\
A= 31 + 2^5+...+2^{10}.31 \\
=> A \vdots 31$

 

[dat20032004@gmail.com]

[tex]\dpi{100} 1+2+2^2+2^3+...+2^{14}\\=(1+2+2^2+2^3+2^4)+(2^5+2^6+2^7+2^8+2^9)+(2^{10}+2^{11}+2^{12}+2^{13}+2^{14})\\=1(1+2+2^2+2^3+2^4)+2^5(1+2+2^2+2^3+2^4)+2^{10}(1+2+2^2+2^3+2^4)\\=(1+2+2^2+2^3+2^4)(1+2^5+2^{10})\\=31.(1+2^5+2^{10})\vdots 31\\ VAY1+2+2^2+2^3+...+2^{14}\vdots 31[/tex]

[tex]\dpi{100} 1+2+2^2+2^3+...+2^{14}\\=(1+2+2^2+2^3+2^4)+(2^5+2^6+2^7+2^8+2^9)+(2^{10}+2^{11}+2^{12}+2^{13}+2^{14})\\=1(1+2+2^2+2^3+2^4)+2^5(1+2+2^2+2^3+2^4)+2^{10}(1+2+2^2+2^3+2^4)\\=(1+2+2^2+2^3+2^4)(1+2^5+2^{10})\\=31.(1+2^5+2^{10})\vdots 31\\ VAY1+2+2^2+2^3+...+2^{14}\vdots 31[/tex]

Cũng được

 

[BhofA]

Ta có: 1+2+2^2+2^3+....+2^12+2^13+2^14
= (1+2+2^2+2^3+2^4) + (2^5+2^6+2^7+2^8+2^9) + (2^10+2^11+2^12+2^13+1^14)
= (1+2+2^2+2^3+2^4) + 2^5.(1+2+2^2+2^3+2^4) + 2^10.(1+2+2^2+2^3+2^4)
= (1+2+2^2+2^3+2^4)(1+2^5+2^10)
= 31.(1+2^5+2^10) chia hết cho 31