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Question

Số tự nhiên n thỏa mãn :$2^0+2^1+2^2+...+2^{21}=2^{2n}-1$

Nguyễn Huy Tú · Accepted Answer

$2^0+2^1+2^2+...+2^{21}=2^{2n}-1$Đặt $A=2^0+2^1+2^2+...+2^{21}=2^{2n}-1$Ta có: $A=2^0+2^1+2^2+...+2^{21}$$\Rightarrow2A=2^1+2^2+2^3+...+2^{22}$$\Rightarrow2A-A=\left(2^1+2^2+2^3+...+2^{22}\right)-\left(2^0+2^1+2^2+...+2^{21}\right)$$\Rightarrow A=2^{22}-2^0$$\Rightarrow A=2^{22}-1$Mà $A=2^{2n}-1$$\Rightarrow2^{22}-1=2^{2n}-1$$\Rightarrow2^{22}=2^{2n}$$\Rightarrow2n=22$$\Rightarrow n=11$Vậy n = 11Câu trả lời: $2^0+2^1+2^2+...+2^{21}=2^{2n}-1$Đặt $A=2^0+2^1+2^2+...+2^{21}=2^{2n}-1$Ta có: $A=2^0+2^1+2^2+...+2^{21}$$\Rightarrow2A=2^1+2^2+2^3+...+2^{22}$$\Rightarrow2A-A=\left(2^1+2^2+2^3+...+2^{22}\right)-\left(2^0+2^1+2^2+...+2^{21}\right)$$\Rightarrow A=2^{22}-2^0$$\Rightarrow A=2^{22}-1$Mà $A=2^{2n}-1$$\Rightarrow2^{22}-1=2^{2n}-1$$\Rightarrow2^{22}=2^{2n}$$\Rightarrow2n=22$$\Rightarrow n=11$Vậy n = 11

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