Question

[Nguyễn Văn Công Hà]

a,\(cho A=2+2^2+2^3+2^4+...+2^9+2^{10}.Chứng minh A⋮3 và A⋮31\\ \)

Answer 1

[Nope...]

Ta có : \(A=2+2^2+2^3+...+2^{10}=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^9+2^{10}\right)\)

\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^9\left(1+2\right)=2.3+2^3.3+...+2^9.3\)

\(=3\left(2+2^3+2^5+...+2^9\right)\) chia hết cho 3

Lại có : \(A=2+2^2+2^3+...+2^{10}=\left(2+2^2+2^3+...+2^5\right)\left(2^6+2^7+2^8+...+2^{10}\right)\)

\(=2\left(1+2+2^2+...+2^4\right)+2^6\left(1+2+2^2+...+2^4\right)=2.31+2^6.31=31\left(2+2^6\right)\) chia hết cho 31

\(\Rightarrow\) A chia hết cho 3 và 31

Answer 2

[Xyz OLM]

a) \(A=2+2^2+2^3+2^4+..+2^9+2^{10}\)

        \(=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^9+2^{10}\right)\)

        \(=\left(2+2^2\right)+2^2.\left(2+2^2\right)+...+2^8.\left(2+2^2\right)\)

        \(=6+2^2.6+...+2^8.6\)  

        \(=6.\left(1+2^2+...+2^8\right)\)

          \(=2.3.\left(1+2^2+..+2^8\right)⋮3\)

Vậy \(A⋮3\left(ĐPCM\right)\)

\(A=2+2^2+2^3+2^4+2^5+2^6+2^7+2^8+2^9+2^{10}\)

\(=\left(2+2^2+2^3+2^4+2^5\right)+\left(2^6+2^7+2^8+2^9+2^{10}\right)\)

\(=\left(2+2^2+2^3+2^4+2^5\right)+2^5.\left(2+2^2+2^3+2^4+2^5\right)\)

\(=62+2^5.62\)

\(=62.\left(1+2^5\right)\)

\(=31.2.\left(1+2^5\right)⋮31\)

Vậy \(A⋮31\left(ĐPCM\right)\)