Question

[le van hoai nam]

Chứng minh 2+2² +2³ +.......+2¹⁰⁰ ckia hết cho 3

Answer 1

[Pham Thi Trang Anh]

gọi A = 2+2^2+2^3+......+2^100

A=(2+2^2+2^3+2^4)+(2^5+2^6+2^7+2^8)+......+(2^97+2^98+2^99+2^100)

A=(2+2^2+2^3+2^4)+(2^5+2^6+2^7+2^8)+......+(2^97+2^98+2^99+2^100)

A=(2+2^2+2^3+2^4).1+(2+2^2+2^3+2^4).4+......+(2+2^2+2^3+2^4).98

A=            30.1          +         30.4             +.......+       30.98

A=                  30.(1+4+...+98)

Vì 30 chia hết cho 3

=>30.(1+4+...98) chia hết cho 3

Hay 2+2^2+2^3+......+2^100 chia hết cho 3

Answer 2

[Nữ Thần Bình Minh]

Gọi \(2+2^2+2^3+...+2^{100}\) là A

\(A=2+2^2+2^3+...+2^{100}\)

\(A=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+...+\left(2^{97}+2^{98}+2^{99}+2^{100}\right)\)

\(A=\left(2+2^2+2^3+2^4\right)1+\left(2+2^2+2^3+2^4\right)4+...+\left(2+2^2+2^3+2^4\right)98\)

\(A=30.1+30.4+...+30.98\)

\(A=30\left(1+4+...+98\right)\)

Vì 30 chia hết cho 3

=> 30 ( 1 + 4 + ... + 98 ) chia hết cho 3

Hay \(2+2^2+2^3+...+2^{100}\) CHIA HẾT CHO 3