Question

Chứng minh tổng sau chia hết cho 7

A  =  2¹ + 2² + 2³ + 2⁴ +....+ 2⁵⁹ + 2⁶⁰

Answer 1

 A=2^1+2^2+...+2^60 
=(2^1+2^2+2^3)+(2^4+2^5+2^6)+(2^7+2^8+2^... 
= ( 2^1+2^2+2^3)*(2^0+2^3+2^6+...+2^57) 
= 14*(2^0+2^3+2^6+...+2^57) chia het cho 7 

ko bt đúng hay sai nx!!

Answer 2

\(A=2^1+2^2+2^3+2^4+...+2^{59}+2^{60}\)

\(\Rightarrow A=\left(2^1+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{58}+2^{59}+2^{60}\right)\)

\(\Rightarrow A=2^1\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{58}\left(1+2+2^2\right)\)

\(\Rightarrow A=2^1\cdot7+2^4\cdot7+...+2^{58}\cdot7\)

\(\Rightarrow A=7\cdot\left(2^1+2^4+...+2^{58}\right)\)

\(\Rightarrow A⋮7\)

Answer 3

Ta có : 

\(A=2^1+2^2+2^3+...+2^{60}\)

\(A=\left(2^1+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{58}+2^{59}+2^{60}\right)\)

\(A=2\left(1+2+4\right)+2^4\left(1+2+4\right)+...+2^{58}\left(1+2+4\right)\)

\(A=2.7+2^4.7+...+2^{58}.7\)

\(A=7\left(2+2^4+...+2^{58}\right)⋮7\)

Vậy \(A⋮7\)

Chúc bạn học tốt ~ 

Answer 4

\(A=2^1+2^2+2^3+2^4+....+2^{59}+2^{60}\)

     \(=\left(2^1+2^2+2^3\right)+...+\left(2^{58}+2^{59}+2^{60}\right)\)

       \(=2\left(1+2+2^2\right)+...+2^{58}\left(1+2+2^2\right)\)

         \(=2.7+...+2^{58}.7\)

          \(=\left(2+...+2^{58}\right).7⋮7\)

\(\Rightarrow A⋮7\)