Question

cho biểu thức \(A=2+2^2+2^3+2^4+...+2^{2014}+2^{2015}+2^{2016}\) 

chứng minh rằng A chia hết cho 7

Answer 1

Ta có:A=(2+2²+2³)+(2⁴+2⁵+2⁶)+..+(2²⁰¹⁴+2²⁰¹⁵+2²⁰¹⁶)

A=2(1+2¹+2²)+2⁴(1+2¹+2²)+...+2²⁰¹⁴(1+2¹+2²)

A=2.7+2⁴.7+...+2²⁰¹⁴.7=7(2+2⁴+...+2²⁰¹⁴)

Suy ra A chia het cho 7

Vậy A chia hết cho 7

Answer 2

\(A=2+2^2+2^3+2^4+...+2^{2014}+2^{2015}+2^{2016}\)

\(\Rightarrow A=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{2014}+2^{2015}+2^{2016}\right)\)

\(\Rightarrow A=2.\left(1+2+2^2\right)+2^4.\left(1+2+2^2\right)+...+2^{2014}.\left(1+2+2^2\right)\)

\(\Rightarrow A=2.7+2^4.7+...+2^{2014}.7\)

\(\Rightarrow A=7.\left(2+2^4+...+2^{2014}\right)\)

\(\Rightarrow A⋮7\)

Answer 3

\(A=2+2^2+2^3+...+2^{2016}\)

\(A=\left(2+2^2+2^3\right)+...+\left(2^{2014}+2^{2015}+2^{2016}\right)\)

\(A=2\left(1+2+2^2\right)+....+2^{2014}\left(1+2+2^2\right)\)

\(A=2\cdot7+...+2^{2014}\cdot7\)

\(A=7\left(2+...+2^{2014}\right)⋮7\)