Question

[Huỳnh Thị Thanh Nhân]

Chứng tỏ tổng :

    \(B=2+2^2+2^3+2^4+....+2^9+2^{10}\) chia hết cho 3 .

Mình cần gấp nên mong các bạn giúp đỡ !

Answer 1

[tth_new]

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

\(2B=2^2+2^3+2^4+...+2^{10}+2^{11}\). Do 2B - B = B nên

\(B=\left(2^2+2^3+2^4+...+2^{10}+2^{11}\right)-\left(2+2^2+2^3+2^4+...+2^9+2^{10}\right)\)

\(=2^{11}-2⋮3^{\left(đpcm\right)}\)

Answer 2

[Phùng Minh Quân]

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

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

\(B=2\left(1+2\right)+2^3\left(1+2\right)+...+2^9\left(1+2\right)\)

\(B=2.3+2^3.3+...+2^9.3\)

\(B=3\left(2+2^3+...+2^9\right)⋮3\) ( đpcm ) 

Vậy \(B⋮3\)