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      

\(A=2+2^2+2^3+2^4+2^5+2^6+2^7+2^8+2^9+2^{10}.\)

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

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

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

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

Vậy \(A⋮3\)

có bạn cộng 2+ 2^2 rồi gộp các số tiếp theo như thế sẽ biết

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

=2.3+2^3.3+2^5.3+2^7.3+2^9.3=3(2+2^3+2^5+2^7+2^9) 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(1+2+2^2)+2^5(1+2+2^2)+2^8(1+2+2^2)=2+7(2^2+2^5+2^8) không chia hết cho 7

=> A không chia hết cho 21 

A = ( 2 + 2² ) + ( 2³ + 2⁴ ) + ( 2⁵ + 2⁶ ) + ( 2⁷ + 2⁸ ) + ( 2⁹ + 2¹⁰ )

A = 2 . 3 + 2³ . 3 + 2⁵ . 3 + 2⁷ . 3 + 2⁹ . 3

A = 3 . ( 2 + 2³ + 2⁵ + 2⁷ + 2⁹ ) \(⋮\) 3

A = 2 + ( 2² + 2³ +2⁴ ) + ( 2⁵ + 2⁶ + 2⁷ ) + ( 2⁸ + 2⁹ + 2¹⁰ )

A = 2 + 2² . ( 1 + 2 + 2² ) + 2⁵ . ( 1 + 2 + 2² ) + 2⁸ . ( 1 + 2 + 2² )

A = 2 + 7 . ( 2² + 2⁵ + 2⁸ ) \(⋮̸\) 7

=> A \(⋮̸\) 21

hok tốt

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

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

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

\(A=3.\left(2+2^3+...+2^9\right)⋮3\)

A = 2 + 2² +...+ 2¹⁰ ( có 10 số hạng)

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

A = 2.(1+2) + 2³.(1+2) + ...+ 2⁹.(1+2)

 A = 2.3 + 2³.3 + ...+ 2⁹.3

A = 3.(2+2³ +...+2⁹) chia hết cho 3

Tổng này CÓ CHIA HẾT CHO 3

Tổng A có: (10-1):1+1=10(số). Ta nhóm như sau:

A=(2+2^2)+(2^3+2^4)+...+(2^9+2^10)

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

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

A=3(2+2^3+...+2^9) chia hết cho 3

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

\(A=\left(2+2^2\right)+\left(2^3+2^4\right)+\left(2^5+2^6\right)+\left(2^7+2^8\right)+\left(2^9+2^{10}\right)\)

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

\(A=2\cdot3+2^3\cdot3+2^5\cdot3+2^7\cdot3+2^9\cdot3\text{ }⋮\text{ }3\)

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

\(A=\left(2+2^2\right)+\left(2^3+2^4\right)+\left(2^5+2^6\right)+\left(2^7+2^8\right)+\left(2^9+2^{10}\right)\)

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

\(A=2\cdot3+2^3\cdot3+2^5\cdot3+2^7\cdot3+2^9\cdot3\)

\(\Rightarrow\text{ }A\text{ }⋮\text{ }3\text{ }\left(\text{ ĐPCM}\right)\)

A= (2+2²)+(2³+2⁴)+...+(2⁹+2¹⁰)

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

A= 2.3+2³.3+...+2⁹.3

A= 3.(2+2³+2⁵+2⁷+2⁹)

Vậy a chia hết cho 3