Question

Cho A = 1 + 2 + 2² + ... + 2¹¹

Không tính tổng A , hãy chứng tỏ A chia hết cho 3

Answer 1

 A = 1 + 2 + 2² + ... + 2¹¹

\(=\left(1+2\right)+\left(2^2+2^3\right)+...+\left(2^{10}+2^{11}\right).\)

\(=3+2^2\left(1+2\right)+2^4\left(1+2\right)+...+2^{10}\left(1+2\right)\)

\(=3\left(1+2^2+2^4+...+2^{10}\right)⋮3\)

Answer 2

A=(1+2)+(2^2+2^3)+...+(2^10+2^11)

  = 3+2^2(1+2)+...+2^10(1+2)

  =3+2^2.3+...+2^10.3

  = 3(1+2^2+...+2^10) chia hết cho 3

=> tổng A chia hết cho 3

Answer 3

Woah