Question

[Nguyễn Thị Anh Thư]

Cho A=1+2+2^2+...+2^2018.

tìm số dư khi chia A cho 10

(nhanh nha,ai nhanh mk tick cho)

Answer 1

[Việt Anh]

A=1+2+2²+...+2²⁰¹⁸

2A=2+2²+2³+...+2²⁰¹⁹

2A-A=(2+2²+2³+...+2²⁰¹⁹) - (1+2+2²+...+2²⁰¹⁸)

A=2²⁰¹⁹ -1

A=2x2²⁰¹⁸-1

A=2²⁰¹⁸:10 dư 2;4;6;8

Answer 2

[Ngô Thế Trường ( CRIS DEVID )]

\(A=1+2+2^2+2^3+.........+2^{2018}\)

\(\Leftrightarrow2A=\left(1+2+2^2+2^3+........+2^{2018}\right).2\)

\(\Leftrightarrow2A=\left(2.1+2.2+2^2.2+2^3.2+........+2^{2018}.2\right)\)

\(\Leftrightarrow2A=2+2^2+2^3+2^4+.........+2^{2019}\)

TRỪ 2 VỀ 2A VỚI A TA CÓ:

\(\Leftrightarrow2A-A=\left(2+2^2+2^3+2^4+......+2^{2019}\right)-\left(1+2+2^2+2^3+.....+2^{2018}\right)\)

\(\Leftrightarrow2A-A=[\left(2-1\right)+\left(2^2-2^2\right)+\left(2^3-2^3\right)+2+.........+2^{2019}]\)

\(\Leftrightarrow A=2^{2019}-1\)

Ta có:\(2^{2019}=2^{2018+1}=2^{2018}.2\Rightarrow A=2^{2018}.2-1+2\)

Vì: \(2^{2018}⋮10\Rightarrow A:10\)Dư 0;2;4;6;8