Câu hỏi của Jungkook Oppa

Question

Tìm m = 2^2010 - ( 2^2009 + 2^2008 + .... + 2^1 + 2^0 )

Minh Anh · Accepted Answer

$M=2^{2010}-\left(2^{2009}+2^{2008}+...+2^1+2^0\right)$$2^{2010}-M=1+2+2^2+...+2^{2008}+2^{2009}$ $2\left(2^{2010}-M\right)=2+2^2+2^3+...+2^{2009}+2^{2010}$$2\left(2^{2010}-M\right)-\left(2^{2010}-M\right)=\left(2+2^2+2^3+...+2^{2009}+2^{2010}\right)-\left(1+2+2^2+...+2^{2008}+2^{2009}\right)$$2^{2010}-M=2^{2010}-1$$M=2^{2010}-2^{2010}+1$$M=1$Câu trả lời: $M=2^{2010}-\left(2^{2009}+2^{2008}+...+2^1+2^0\right)$$2^{2010}-M=1+2+2^2+...+2^{2008}+2^{2009}$ $2\left(2^{2010}-M\right)=2+2^2+2^3+...+2^{2009}+2^{2010}$$2\left(2^{2010}-M\right)-\left(2^{2010}-M\right)=\left(2+2^2+2^3+...+2^{2009}+2^{2010}\right)-\left(1+2+2^2+...+2^{2008}+2^{2009}\right)$$2^{2010}-M=2^{2010}-1$$M=2^{2010}-2^{2010}+1$$M=1$

Nguyễn Huy Tú · Answer

Đặt $M=2^{2010}-A$Ta có:$A=2^{2009}+2^{2008}+...+2^1+2^0$$\Rightarrow2A=2^{2010}+2^{2009}+...+2^2+2^1$$\Rightarrow2A-A=\left(2^{2010}+2^{2009}+...+2^2+2^1\right)-\left(2^{2009}+2^{2008}+...+2^1+2^0\right)$$\Rightarrow A=2^{2010}-1$$\Rightarrow M=2^{2010}-\left(2^{2010}-1\right)$$\Rightarrow M=\left(2^{2010}-2^{2010}\right)+1$$\Rightarrow M=1$Câu trả lời: Đặt $M=2^{2010}-A$Ta có:$A=2^{2009}+2^{2008}+...+2^1+2^0$$\Rightarrow2A=2^{2010}+2^{2009}+...+2^2+2^1$$\Rightarrow2A-A=\left(2^{2010}+2^{2009}+...+2^2+2^1\right)-\left(2^{2009}+2^{2008}+...+2^1+2^0\right)$$\Rightarrow A=2^{2010}-1$$\Rightarrow M=2^{2010}-\left(2^{2010}-1\right)$$\Rightarrow M=\left(2^{2010}-2^{2010}\right)+1$$\Rightarrow M=1$

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