Question

Bài 4.  Cho A = 2⁰ + 2¹ + 2² + 2³ + .... + 2⁵⁹.

Chứng tỏ A chia hết cho 7

Answer 1

giúp mik với 

Answer 2

\(A=2^0+2^1+2^2+...+2^{59}\)

\(=2^0\left(1+2+2^2\right)+2^3\left(1+2+2^2\right)+...+2^{57}\left(1+2+2^2\right)\)

\(=2^0.7+2^3.7+...+2^{57}.7\)

\(=7\left(2^0+2^3+...+2^{57}\right)⋮7\)