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

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

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

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

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

⇒ \(A=B\)

A = 1 + 2 + 2² + ... + 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

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

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

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

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

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

A = 1 + 2 + 2² + ... + 2²⁰⁰² 

=> 2A = 2(1 + 2 + 2² + ... + 2²⁰⁰² )

= 2 + 2² + 2³ + .... + 2²⁰⁰³

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

A = 2²⁰⁰³ - 1

Vì 2²⁰⁰³ - 1 < 2²⁰⁰³ nên A < B

Vậy A < B

A < B

Ủng hộ nha

Cảm ơn nhiều

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

=> \(A=2A-A=2^{2003}-1< B\)

2/

A=1+2+2^2+...+2^10

2.A= 2+2^2+...+2^11

=>2A-A = 2^11-1=> A = 2^11 -1=B

Vậy A=B

1)5²⁰⁰³+5²⁰⁰²+5²⁰⁰¹=5²⁰⁰¹(5²+5+1)=5²⁰⁰¹(25+5+1)=5²⁰⁰¹.31

Vì 31 chia hết cho 31nên

5²⁰⁰¹.31chia hết cho 31 hay 5²⁰⁰³+5²⁰⁰²+5²⁰⁰¹ chia hết cho 31

2) A = 1+2+2²+......+2⁹+2¹⁰

=>2A=2+2²+2³+...+2¹¹

=>2A-A=2+2²+2³+...+2¹¹-(1+2+2²+...+2⁹+2¹⁰)

=>A=2¹¹-1

Vậy A=B=2¹¹-1

a ) Ta có:

A = 1 + 3 + 3² + 3³+ ..... + 3⁶

A x 3 = 3 + 3² + 3³ + 3⁴ + .... + 3⁷

A x 2 - A = ( 3 + 3² + 3³ + 3⁴ + .... + 3⁷ ) - ( 1 + 3 + 3² + 3³ + .... + 3⁶ )

A = 3⁷ - 1

Mà : B = 3⁷ - 1 nên A = B

b ) Ta có :

C = 1 + 2 + 2² + 2³ + ...... + 2²⁰⁰²

C x 2 = 2 + 2² + 2³ + 2⁴ + ..... + 2²⁰⁰³

C x 2 - C = ( 2 + 2² + 2³ + 2⁴ + ...... + 2²⁰⁰³ ) - ( 1 + 2 + 2² + 2³ + ..... + 2²⁰⁰² )

C = 2²⁰⁰³ - 1 

Mà : D = 2²⁰⁰³ - 1 nên C = D

A=1+3+3^2+....+3^100
\Rightarrow 3A=3+ + +...+ 
\Rightarrow3A-A=2A=(3+ + + )-(1+3+ +....+ )
= -1
\RightarrowA=( -1):2