Question

A = 4+\(2^2+2^3+2^4+.......+2^{20}\)

Hỏi A có chia hết cho 128 ko ?

giúp mình với nha

Answer 1

A = 4+2^2+2^3+2^4+.......+2^20

A=2^2+2^2+2^3+2^4+.......+2^20

2A=2.(4+2^2+2^3+2^4+.......+2^20)

2A=2^3+2^3+2^4+2^5+...+2^21

2A-A=(2^3+2^3+2^4+2^5+...+2^21-(2^2+2^3+2^4+.......+2^20)

A=2^3+2^21-(2^2+2^2)

A=8+2^21-8

A=2^21

-Ta có:128=2^7

Mà 2^21 chia hết cho 2^7

Vậy A chia hết cho 128.

Nhớ tick nha!

Answer 2

A=4+2²+2³+2⁴+...+2²⁰

=>2.A=2³+2⁴+2⁵+...+2²¹

=>2A-A=2³+2⁴+2⁵+...+2²¹-(2²)-(2²)-(2³)-(2⁴)-...-(2²⁰)

=>A= -(2²)-(2²)

=>A= -8

Vậy A= -8

Answer 3

A=4+2\(^2\)+2³+2⁴+...+2²⁰

=>2A=2\(^3\)+2\(^3\)+2⁴+2⁵+...+2²¹

=>2A-A=2\(^3\)+2\(^3\)+2⁴+2⁵+...+2²¹-(2²)-(2\(^2\))-(2³)-(2⁴)-...-(2²⁰)

=>A=2\(^3\)+2\(^{21}\)-(2²)-(2\(^2\))

=>A=8+2²¹-8

=>A=2⁷.2¹⁴ chia hết cho 128

Vậy A chia hết cho 128