Câu hỏi của Lê Hoàng Hồng Mai 6A5

Question

Lấp La Lấp Lánh · Accepted Answer

$A=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{2009}\left(1+2\right)=2.3+2^3.3+...+2^{2009}.3=3\left(2+2^3+....+2^{2009}\right)⋮3$Câu trả lời: $A=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{2009}\left(1+2\right)=2.3+2^3.3+...+2^{2009}.3=3\left(2+2^3+....+2^{2009}\right)⋮3$

ILoveMath · Answer

$A=2^1+2^2+2^3+2^4+...+2^{2010}$$\Rightarrow A=\left(2^1+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{2009}+2^{2010}\right)$$\Rightarrow A=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{2009}\left(1+2\right)$$\Rightarrow A=\left(1+2\right)\left(2+2^3+...+2^{2009}\right)$$\Rightarrow A=3\left(2+2^3+...+2^{2009}\right)3⋮$Câu trả lời: $A=2^1+2^2+2^3+2^4+...+2^{2010}$$\Rightarrow A=\left(2^1+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{2009}+2^{2010}\right)$$\Rightarrow A=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{2009}\left(1+2\right)$$\Rightarrow A=\left(1+2\right)\left(2+2^3+...+2^{2009}\right)$$\Rightarrow A=3\left(2+2^3+...+2^{2009}\right)3⋮$

Shinichi Kudo · Answer

$A=2^1+2^2+2^3+...+2^{2010}$$A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{2009}+2^{2010}\right)$$A=\left(2+2^2\right)+2^2\left(2+2^2\right)+...+2^{2008}\left(2+2^2\right)$$A=\left(2+2^2\right)\left(1+2^2+...+2^{2008}\right)$mà $2+2^2⋮3$ $\Rightarrow\left(2+2^2\right)\left(1+2^2+...+2^{2008}\right)⋮3$hay $A=2^1+2^2+2^3+...+2^{2010}$ chia hết cho 3Câu trả lời: $A=2^1+2^2+2^3+...+2^{2010}$$A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{2009}+2^{2010}\right)$$A=\left(2+2^2\right)+2^2\left(2+2^2\right)+...+2^{2008}\left(2+2^2\right)$$A=\left(2+2^2\right)\left(1+2^2+...+2^{2008}\right)$mà $2+2^2⋮3$ $\Rightarrow\left(2+2^2\right)\left(1+2^2+...+2^{2008}\right)⋮3$hay $A=2^1+2^2+2^3+...+2^{2010}$ chia hết cho 3

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