Question

a.Chứng minh rằng tổng sau có chia hết cho 3 không?

A=2+2²+2³+2⁴+2⁵+2⁶+2⁷+2⁸+2⁹+2¹⁰

 

Answer 1

A=2+2²+2³+2⁴+2⁵+2⁶+2⁷+2⁸+2⁹+2¹⁰

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

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

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

\(=\left(2+2^3+...+2^{10}\right).3\) chia hết cho 3

Answer 2

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

      A=2.(1+2)+2³.(1+2)+2⁵.(1+2)+2⁷.(1+2).2⁷.(1+2)+2⁹.1+2

      A=2.3+2³.3+2⁵.3+2⁷.3+2⁹.3

Vậy A chia hết cho 3

Answer 3

taco A=(2+2^2)+...+(2^9+2^10)

A=3.2+...+3.2^9

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

Answer 4

Vây A là 3 nha Kuga Shu

Answer 5

A=(2+2²)+(2³+2⁴)+....+(2⁹+2¹⁰)

A=(1+2).2+(1+2).8+....+(1+2).512

A=3.2+3.8+...+3.512

A=3(2+8+...+512) CHIA HẾT CHO 3 

=> A CHIA HET CHO 3

Answer 6

2+2^2+2^3+...+2^10=(2+2^2)+(2^3+2^4)+...+(2^9+2^10)=(2+2^2)+2^2*(2+2^2)+...+2^8*(2+2^2)=6+2^2*6+...+2^8*6

còn tiếp theo tự làm

Answer 7

có

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

A=6+2^2.6+...+2^8.6

A=6.(1+4+...+2^8)chia hết cho 3

Answer 8

A=2(1+2)+\(2^3\)(1+2)+...+\(2^9\)(1+2)=2.3+8.3+...+512.3=(2+8+...+512)3 chia hết cho 3 đpcm