Câu hỏi của Đỗ Nhật Minh

Question

A=1+2^1+2^2+...+2^100+2^101.Chứng tỏ rằng A chia hết cho 7

Thịnh Nguyễn Công · Accepted Answer

Ta có:A=1+21+22+...+2100+2101A=1+21+22+...+2100+2101= (1+2+22)+(23+24+25)+...+(299+2100+2101)(1+2+22)+(23+24+25)+...+(299+2100+2101)= (1+2+22)+22.(1+2+22)+...+299.(1+2+22)(1+2+22)+22.(1+2+22)+...+299.(1+2+22)= (1+2+22).(1+22+26+...+299)(1+2+22).(1+22+26+...+299)= 7.(1+22+26+...+299)⋮77.(1+22+26+...+299)⋮7(Vì 7⋮7)                      Câu trả lời: Ta có:A=1+21+22+...+2100+2101A=1+21+22+...+2100+2101= (1+2+22)+(23+24+25)+...+(299+2100+2101)(1+2+22)+(23+24+25)+...+(299+2100+2101)= (1+2+22)+22.(1+2+22)+...+299.(1+2+22)(1+2+22)+22.(1+2+22)+...+299.(1+2+22)= (1+2+22).(1+22+26+...+299)(1+2+22).(1+22+26+...+299)= 7.(1+22+26+...+299)⋮77.(1+22+26+...+299)⋮7(Vì 7⋮7)

ILoveMath · Answer

$A=1+2^1+2^2+...+2^{100}+2^{101}$$\Rightarrow A=\left(1+2^1+2^2\right)+\left(2^3+2^4+2^5\right)+...+\left(2^{99}+2^{100}+2^{101}\right)$$\Rightarrow A=\left(1+2^1+2^2\right)+2^3\left(1+2^1+2^2\right)+...+2^{99}\left(1+2^1+2^2\right)$$\Rightarrow A=\left(1+2^1+2^2\right)\left(1+2^3+...+2^{99}\right)$$\Rightarrow A=7\left(1+2^3+...+2^{99}\right)⋮7$Câu trả lời: $A=1+2^1+2^2+...+2^{100}+2^{101}$$\Rightarrow A=\left(1+2^1+2^2\right)+\left(2^3+2^4+2^5\right)+...+\left(2^{99}+2^{100}+2^{101}\right)$$\Rightarrow A=\left(1+2^1+2^2\right)+2^3\left(1+2^1+2^2\right)+...+2^{99}\left(1+2^1+2^2\right)$$\Rightarrow A=\left(1+2^1+2^2\right)\left(1+2^3+...+2^{99}\right)$$\Rightarrow A=7\left(1+2^3+...+2^{99}\right)⋮7$

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