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

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

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

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

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

⇒ \(A=B\)

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

ta có : a = 1 + 2 + 2^2 + ... + 2^2002

=> 2a = 2 + 2^2 + 2^3 + ... + 2^2003

=> 2a-a = (2+2^2 + 2^3 + ... + 2^2003) - ( 1+2+2^2+...+2^2002)

=> a = 2^2003 - 1

Vậy 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

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

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

\(\Rightarrow2A-A=2^{2013}-1=B\)

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

Đặt d=UCLN(12n+1, 30n+1)

Khi đó: \(\hept{\begin{cases}12n+1⋮d\\30n+1⋮d\end{cases}}\)<=> 5(12n+1)-2(30n+1)\(⋮\)d <=> 3\(⋮\)d

Nên d=1 hoặc d=3

Mặt khác: 12n\(⋮\)3=> 12n+1 không chia hết cho 3

do đó d\(\ne\)3

Vậy d=1 (ĐPCM)

2, 

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

=> \(A=2A-A=2^{2003}-1< B\)

Ta có:\(A=1+2+2^2+2^3+...+2^{2002}\)

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

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

\(2A-B=\left(2+2^2+2^3+2^4+...+2^{2003}\right)-2^{2003}\)

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

\(\Rightarrow A-1=2A-B\)

\(\Rightarrow A>B\)

so sánh 1 và 2 là biết