vì các mũ trên ko chia hết cho 3 nên tổng trên ko chia hết cho 3

A chia hết cho 3

100 %

A = ( 2 + 2^2 ) + ( 2^3 + 2^4 ) + ( 2^5 + 2^6 ) + ( 2^7 + 2^8 ) + ( 2^9 + 2^10 )

A = ( 2 . 1 + 2 . 2 ) + ( 2^3 .1 + 2^3 .2  ) + ( 2^5 . 1 + 2^5 . 2 ) + ( 2^7 . 1 + 2^7 . 2 ) + ( 2^9 . 1 + 2^9 . 2 ) 

A = 2( 1 + 2 ) + 2^3( 1 + 2 ) + 2^5( 1 + 2 ) + 2^7( 1 + 2 ) + 2^9( 1 + 2 )

A = 2.3 + 2^3.3 + 2^5.3 + 2^7.3 + 2^9.3

A = 3( 2 + 2^3 + 2^5 + 2^7 + 2^9 )

=> A chia hết cho 3 

A=2+2^2+2^3+2^4+2^5+2^6+2^7+2^8+2^9+2^10

=(2+2^2)+(2^3+2^4)+(2^5+2^6)+(2^7+2^8)+(2^9+2^10)

=2(1+2)+2^3(1+2)+2^5(1+2)+2^7(1+2)+2^9(1+2)

=(1+2)(2+2^3+2^5+2^7+2^9) 

=3(2+2^3+2^5+2^7+2^9) chia hết cho 3

ko thì phải

\(=2\left(1+2\right)+2^3\left(1+2\right)+2^5\left(1+2\right)+2^7\left(1+2\right)+2^9\left(1+2\right)\)

\(=3\left(2+2^3+2^5+2^7+2^9\right)⋮3\)

Ta có :

\(2+2^2+2^3+....+2^{10}\)

\(=2\left(1+2\right)+2^3\left(1+2\right)+....+2^9\left(1+2\right)\)

\(=2.3+2^3.3+....+2^9.3\)

=> Tổng chia hết cho 3

\(2+2^2+2^3+...+2^{10}\)

\(=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^9+2^{10}\right)\)

\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^9\left(1+2\right)\)

\(=2.3+2^3.3+...+2^9.3\)

\(=\left(2+2^3+...+2^9\right).3⋮3\)

\(\Rightarrow2+2^2+2^3+...+2^{10}⋮3\)

có

ta thấy

\(2+2^2=4\\->2+2^2+...+2^{10}\) chia hết cho 3

Nguyễn Huy Hải ns chuyện vs gái "'hiền"' gê nhể ! 

A=(2+2^2)+(2^3+2^4)+(2^5+2^6)+(2^7+2^8)+(2^9+2^10)

A=(2.1+2.2)+...+(2^9.1+2^9.2)

A=2.3+2^3.3+...+2^9.3

A=3.(2+2^3+...+2^9) chia hết cho 3

=> A chia hết cho 3

A => (2 + 2²) + (2³ + 2⁴) +.... + (2⁹ + 2¹⁰)

A => 2.3 + 2³.3 + .... + 2⁹.3

A => 3.(2+2³+2⁵ + 2⁷+2⁹)

Vậy A có thể chia hết cho 3