Ta có : \(A=1+2+2^2+...+2^{2017}\)(1)

\(\Rightarrow2A=2+2^2+2^3+...+2^{2018}\)(2)

Lấy (2) trừ (1) ta có : 

\(\Rightarrow A=2^{2018}-1\)

\(\Rightarrow A< B\). Vì \(B=2^{2018}\)

A = 1+2+2²+2³+.....+2²⁰¹⁷

2A = 2(1+2+2²+2³+.....+2²⁰¹⁷)  = 2+2²+2³+2⁴+.....+2²⁰¹⁸

2A - A = 2+2²+2³+2⁴+.....+2²⁰¹⁸- (1+2+2²+2³+.....+2²⁰¹⁷)

=> A = 2+2²+2³+2⁴+.....+2²⁰¹⁸-1-2-2²-2³-.....-2²⁰¹⁷

       A =2²⁰¹⁸-1 < 2²⁰¹⁸

Vậy A < B

Ta có: \(A=1+2+2^2+...+2^{2017}\)

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

\(2A-A=2+2^2+2^3+...+2^{2018}-\left(1+2+2^2+...+2^{2017}\right)\)

\(A=2^{2018}-1\)

\(\Rightarrow A-B=2^{2018}-1-2^{2018}=-1\)

A=1+2+2²+2³+...+2^{2017 (1)}

2A=2+2²+2³+2⁴+...+2^{2018 (2)}

Lấy (2) - (1) ta có:

2A - A=(2+2²+2³+2⁴+...+2²⁰¹⁸)-(1+2+2²+2³+...+2²⁰¹⁷)

A=2+2²+2³+2⁴+...+2²⁰¹⁸-1-2-2²-2³-...-2²⁰¹⁷

A=2²⁰¹⁸-1

Mà B=2²⁰¹⁸-1 =>A=B

b) ta có: B=2017²

B=(2016+1).2017=2016.2017+2017

A=2016.2018

A=2016.(2017+1)=2016.2017+2016

Vì 2016<2017=>A<B

mình nhé

a, A = 1+2+2²+...+2²⁰¹⁷

2A=2+2²+2³+...+2²⁰¹⁸

2A-A=A=2²⁰¹⁸-1

=> A  = B

b, A = 2016.2018 =2016.(2017+1)=2016+2017.2016

B=2017²=2017.2017=2017.(2016+1)=2017.2016+2017

Vì 2016 < 2017 => 2016+2017.2016 < 2017.2016+2017 => A < B

cảm ơn nhé, Hoàng

bạn viết lại đề đc ko bạn:>,ko hỉu đề

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Cộng à bn

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

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

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

\(A=1+4+4^2+...+4^{2017}\)

=>\(4\cdot A=4+4^2+4^3+...+4^{2018}\)

=>\(4A-A=4+4^2+...+4^{2018}-1-4-4^2-...-4^{2017}\)

=>\(3A=4^{2018}-1\)

=>\(A=\dfrac{4^{2018}-1}{3}\)

\(2B-A=\dfrac{4^{2018}}{6}\cdot2-\dfrac{4^{2018}-1}{3}\)

\(=\dfrac{4^{2018}}{3}-\dfrac{4^{2018}-1}{3}=\dfrac{1}{3}\)