Question

Cho A =2+2²+2³+.....+2²⁰²⁰+2²⁰²¹+2²⁰²²

^{CHỨNG TỎ rằng A chia hết cho3}

Answer 1

\(A=2+2^2+2^3+...+2^{2020}+2^{2021}+2^{2022}\\=(2+2^2)+(2^3+2^4)+(2^5+2^6)+...+(2^{2021}+2^{2022})\\=2\cdot(1+2)+2^3\cdot(1+2)+2^5\cdot(1+2)+...+2^{2021}\cdot(1+2)\\=2\cdot3+2^3\cdot3+2^5\cdot3+...+2^{2021}\cdot3\\=3\cdot(2+2^3+2^5+..+2^{2021})\)

Vì \(3\cdot\left(2+2^3+2^5+...+2^{2021}\right)⋮3\)

nên \(A⋮3\).

\(Toru\)

Answer 2

A=(2+2²)+2²(2+2²)+...+2²⁰²⁰(2+2²)

A= 6.1+2².6+...+2²⁰²⁰.6

A=6(1+2²+...+2²⁰²⁰) chia hết cho 3

vậy A chia hết cho 3

Answer 3

A=(2+2²)+(2³+2⁴)+(2⁵+2⁶)+.......+(2²⁰¹⁹+2²⁰²⁰)+(2²⁰²¹+2²⁰²²)

A=2.(1+2)+2³.(1+2)+2⁵.(1+2)+.......+2²⁰¹⁹.(1+2)+2²⁰²¹.(1+2)

A=2.3+2³.3+2⁵.3+.......+2²⁰¹⁹.3+2²⁰²¹.3

A=3.(2+2³+2⁵+........+2²⁰¹⁹+2²⁰²¹)

Vì 3⋮3⇒A⋮3