Question

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

Answer 1

Ta có:

 

 

 

 

(Vì )                     

 

Answer 2

\(A=1+2^1+2^2+...+2^{100}+2^{101}\)

\(\Rightarrow A=\left(1+2^1+2^2\right)+\left(2^3+2^4+2^5\right)+...+\left(2^{99}+2^{100}+2^{101}\right)\)

\(\Rightarrow A=\left(1+2^1+2^2\right)+2^3\left(1+2^1+2^2\right)+...+2^{99}\left(1+2^1+2^2\right)\)

\(\Rightarrow A=\left(1+2^1+2^2\right)\left(1+2^3+...+2^{99}\right)\)

\(\Rightarrow A=7\left(1+2^3+...+2^{99}\right)⋮7\)