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

\(\Leftrightarrow2A=2+2^2+....+2^{2018}+2^{2019}\)

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

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

Vậy \(A< 2^{2019}\)

bạn viết lại đề đc ko bạn:>,ko hỉu đề

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Cộng à bn

2/3A=2/3-(2/3)^2+...+(2/3)^2019-(2/3)^2020

=>5/3A=1-(2/3)^2020

=>A=(3^2020-2^2020)/3^2020:5/3=\(\dfrac{3^{2020}-2^{2020}}{3^{2020}}\cdot\dfrac{3}{5}=\dfrac{3^{2020}-2^{2020}}{5\cdot3^{2019}}\) ko là số nguyên

Ta có : \(A=2^0+2^1+2^2+...+2^{2018}\)

\(\Rightarrow2A=2^1+2^2+2^3+...+2^{2019}\)

\(\Rightarrow2A-A=2^{2019}-2^0\)

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

\(\Rightarrow A=B\)

Tham khảo nak ~

\(S=2^{2019}-2^{2018}-2^{2017}-...-2^2-2-1\)

   \(=2^{2019}-\left(1+2+2^2+...+2^{2017}+2^{2018}\right)\) (1)

Đặt \(Q=1+2+2^2+...+2^{2017}+2^{2018}\)

\(2Q=2+2^2+2^3+...+2^{2018}+2^{2019}\)

\(2Q-Q=2^{2019}-1\)

\(Q=2^{2019}-1\)(2) 

Từ (1) và (2), ta được:

\(S=2^{2019}-\left(2^{2019}-1\right)=1\)

\(3^{202}:3^{199}-4^{301}.4^{199}\)

\(=3^{202-199}-4^{301+199}\)

\(=3^3-4^{500}\)

\(=9-4^{500}\)

S = 1-3 + 3² - 3³ + ..+ 3²⁰¹⁸ - 3²⁰¹⁹

=> 3S = 3 - 3² + 3³ - 3⁴ +...+ 3²⁰¹⁹ - 3²⁰²⁰

=> 3S + S = 1 - 3²⁰²⁰

4S = 1 - 3²⁰²⁰

\(S=\frac{1-3^{2020}}{4}\)

bài này phải là 1 + đó bạn