\(A=1+2+2^2+2^3+...+2^{2021}\)

\(\Rightarrow2A=2+2^2+2^3+...+2^{2022}\)

\(\Rightarrow A=2A-A=2+2^2+...+2^{2022}-1-2-2^2-...-2^{2021}=2^{2022}-1>2^{2021}-1=N\)

\(a=1+2+2^2+...+2^{2021}\\ \Rightarrow2a=2+2^2+2^3+...+2^{2022}\\ \Rightarrow2a-a=\left(2+2^2+2^3+...+2^{2022}\right)-\left(1+2+2^2+...+2^{2021}\right)\\ \Rightarrow a=2^{2022}-1>2^{2021}-1=n\)

giúp mk vs

\(a=1+2+2^2+...+2^{2021}\)

\(\Rightarrow2a=2+2^2+2^3+...+2^{2022}\)

\(\Rightarrow2a-a=2+2^2+2^3+...+2^{2022}-1-2-2^2-...-2^{2021}\)

\(\Rightarrow a=2^{2022}-1\)

\(\Rightarrow a=2^{2022}-1=b\)

\(a=1+2+2^2+2^3+...+2^{2021}\)

\(2a=2+2^2+2^3+2^4...+2^{2021}+2^{2022}\)

\(2a-a=\)\(\left(2+2^2+2^3+2^4...+2^{2021}+2^{2022}\right)-\left(1+2+2^2+2^3+...+2^{2021}\right)\)

\(a=2^{2022}-1\)

⇒ a=b

\(\Rightarrow2A=2+2^2+2^3+...+2^{2003}\\ \Rightarrow2A-A=2+2^2+2^3+...+2^{2003}-1-2-...-2^{2002}\\ \Rightarrow A=2^{2003}-1=B\)

\(A=1+2+2^2+...+2^{2002}\)

\(2A=2+2^2+2^3+...+2^{2003}\)

\(2A-A=\left(2+2^2+2^3+...+2^{2003}\right)-\left(1+2+2^2+...+2^{2002}\right)\)

\(A=2^{2003}-1\)

⇒ \(A=B\)

A = 1 + 2 + 2² + 2³ + ... + 2²⁰⁰²

=> 2A = 2 + 2² + 2³ + 2⁴ + ... + 2²⁰⁰³

=> 2A - A = ( 2 + 2² + 2³ + 2⁴ + ... + 2²⁰⁰³ ) - ( 1 + 2 + 2² + 2³ + ... + 2²⁰⁰² )

A = 2²⁰⁰³ - 1 < 2²⁰⁰³ 

hay A < B

Vậy ...

\(A=1+2+2^2+2^3+...+2^{2002}\)

\(\Rightarrow2A=2+2^2+2^3+...+2^{2002}+2^{2003}\)

\(\Rightarrow2A-A=2^{2003}-1\)

\(\Rightarrow A=2^{2003}-1\)

Vì \(2^{2003}-1< 2^{2003}\)

nên A < B

\(A=1+3+3^2+...+3^{2001}\)

\(\Rightarrow3A=3+3^2+3^3+...+3^{2002}\)

\(\Rightarrow3A-A=3+3^2+3^3+...+3^{2002}-1-3^2-3^3-...-3^{2001}\)

\(\Rightarrow2A=3^{2002}-1\)

\(\Rightarrow A=\dfrac{3^{2002}-1}{2}\)

Vì \(\dfrac{3^{2002}-1}{2}< 3^{2002}-1\Rightarrow A< B\)

A = 1 + 2 + 2² + ... + 2^2002  

A = 1 + (2 + 2² + ... + 2^2002 )  

Ta xét :  

u1 = 2  

u2 = 2.2 = 2²  

u3 = 2.2² = 2^3  

u2002 = 2.2^2001 = 2^2002  

Tổng cấp số nhân : S = u1.(1 - q^n) / (1 - q) = 2.(1 - 2^2002) / (1 - 2) = 2(2^2002 - 1) = 2^2003 - 2  

A = 1 + 2^2003 - 2 = 2^2003 - 1  

So sánh với B  

2^2003 - 1 = 2^2003 - 1

 Vậy B = A 

A<B                      

=>2A=2+2^2+2^3+2^4+2^5+...+2^2002+2^2003

=>2A-A=2^2003-1

=>A=2^2003-1

=>A<B

giúp mình với

nhanh nhanh nhanh nhanh nhanh nhanh nhanh nhanh

\(A=1+2+2^2+...+2^{2020}\)

\(\Rightarrow2A=2+2^2+2^3+...+2^{2021}\)

\(\Rightarrow2A-A=2+2^2+2^3+...+2^{2021}-1-2-2^2-...-2^{2020}\)

\(\Rightarrow A=2^{2021}-1\)

\(\Rightarrow A=2^{2021}-1=B\)

sorry mình chưa học

Ta có:

\(B=2^{2012}+2^{2011}+...+2^3+2^2+2+1\)

\(\Rightarrow2B=2^{2013}+2^{2012}+...+2^4+2^3+2^2+2\)

\(\Rightarrow2B-B=\left(2^{2013}+2^{2012}+...+2^4+2^3+2^2+2\right)-\left(2^{2012}+...+1\right)\)

\(\Rightarrow B=2^{2013}-1\)

\(A=2^{2003}.9+2^{2003}.1005\)

\(\Rightarrow A=2^{2003}.\left(9+1005\right)\)

\(\Rightarrow A=2^{2003}.1024\)

\(\Rightarrow A=2^{2003}.2^{10}\)

\(\Rightarrow A=2^{2013}\)

Vì \(2^{2013}-1< 2^{2013}\) nên A > B

Vậy A > B