\(A=1+2+2^2+2^3+............+2^{11}\)

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

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

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

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

=>đpcm

A=(1+2)+(2²+2³)+...+(2¹⁰+2¹¹)

A=3+2².(1+2)+...+2¹⁰.(1+2)

A=3+2².3+...+2¹⁰.3

A=3+(2²+...+2¹⁰)

=>A:cho 3

tick mk nha

 Làm  thế  nào đây ?

ai giúp mình với

Ko bt

 A = 1 + 2 + 2² + ... + 2¹¹

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

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

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

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

Woah

cứ tổng hai số hạng sẽ chia hết cho 3 nhé 

\(A=1+2+2^2+2^3+...+2^{11}\)

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

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

\(A=3+2^2.3+...+2^{10}.3\)

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

\(\Rightarrow A⋮3\)                  

Vậy \(A⋮3\)

  !!!

ok mình làm đc r thank you bạn nha@

2A=2+2^2+2^3+...+2^12

2A-A=(2+2^2+2^3+...+2^12)-(1+2+2^2+2^3+...+2^11)

A=2^12-1

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

A=3+2^2.3+...+2^10.3

A=3.(1+2^2+2^4+...+2^10)chia hết cho 3

A=(1+2+2^2)+...+(2^9+2^10+2^11)

A=7+7.2^3+...+2^9.7

A=7(1+2^3+...+2^9)chia hết cho 7

A=(1+2)+(2²+2³)+...+(2¹⁰+2¹¹)

A=3+2²(1+2)+...+2¹⁰(1+2)

A=3+2².3+...+2¹⁰.3

A=3(1+2²+...+2¹⁰)chia hết cho 3

=>1+2 +2²+2³+....+2¹¹ chia hết cho 3