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Question

chứng tỏ tổng A =2+22+23 + 24+.....+28+29+ 210 chia hết cho 3

Phạm Tuấn Đạt · Accepted Answer

Ta có :$A=2+2^2+2^3+2^4+...+2^9+2^{10}$$\Rightarrow A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^9+2^{10}\right)$$\Rightarrow A=2\left(1+2\right)+2^3\left(1+2\right)+...+2^9\left(1+2\right)$$\Rightarrow A=2.3+2^3.3+...+2^9.3$$\Rightarrow A=3\left(2+2^2+...+2^9\right)$$\Rightarrow A⋮3$$\RightarrowĐPCM$Câu trả lời: Ta có :$A=2+2^2+2^3+2^4+...+2^9+2^{10}$$\Rightarrow A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^9+2^{10}\right)$$\Rightarrow A=2\left(1+2\right)+2^3\left(1+2\right)+...+2^9\left(1+2\right)$$\Rightarrow A=2.3+2^3.3+...+2^9.3$$\Rightarrow A=3\left(2+2^2+...+2^9\right)$$\Rightarrow A⋮3$$\RightarrowĐPCM$

My Nguyễn Thị Trà · Answer

$A=2+2^2+2^3+...+2^{10}$$A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^9+2^{10}\right)$$A=2.\left(1+2\right)+2^3.\left(1+2\right)+.....+2^9.\left(1+2\right)$$A=2.3+2^3.3+....+2^9.3$$A=3.\left(2+2^3+....+2^9\right)$$\Leftrightarrow A⋮3$Vậy $A⋮3$Câu trả lời: $A=2+2^2+2^3+...+2^{10}$$A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^9+2^{10}\right)$$A=2.\left(1+2\right)+2^3.\left(1+2\right)+.....+2^9.\left(1+2\right)$$A=2.3+2^3.3+....+2^9.3$$A=3.\left(2+2^3+....+2^9\right)$$\Leftrightarrow A⋮3$Vậy $A⋮3$

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