A = 1 + 3 + 3² + ... + 3²⁰²³

⇒ 3A = 3 + 3² + 3³ + ... + 3²⁰²³ + 3²⁰²⁴

⇒ 2A = 3A - A

= (3 + 3² + 3³ + ... + 3²⁰²³ + 3²⁰²⁴) - (1 + 3 + 3² + ... + 3²⁰²³)

= 3²⁰²⁴ - 1

⇒ A = (3²⁰²⁴ - 1) : 2

⇒ A < B

A=1+3+3²+3³+3⁴+........+3²⁰²²+3²⁰²³

3A=3+3²+3³+............+3²⁰²³+3²⁰²⁴

3A-A=(3+3²+3³+..........+3²⁰²³+3²⁰²⁴

A=1+3+3²+...+3²⁰²²+3²⁰²³

3A=3+3²+3³+...+3²⁰²³+3²⁰²⁴

3A-A=(3+3²+3³+...+3²⁰²³+3²⁰²⁴)-(1+3+3²+...+3²⁰²²+3²⁰²³)

2A=3²⁰²⁴-1

A=(3²⁰²⁴-1):2

ta thấy 3²⁰²⁴-1 lớn hơn (3²⁰²⁴-1):2

vậy B lớn A

anh tai sadboy

Trường nào đó?

A =1+3+3²+.....+3²⁰²²+3²⁰²³

3.A =3+3²+3³+.....+3²⁰²³+3²⁰²⁴

3.A -A=(3+3²+3³+.....+3²⁰²³+3²⁰²⁴ ) - (1+3+3²+.....+3²⁰²²+3²⁰²³)

2A =3²⁰²⁴-1

A =\(\dfrac{3^{2024}-1}{2}\)

       A = 1 + 3 + 3² + 3³ + 3⁴ + ... + 3²⁰²²

     3A = 3  + 3² + 3³ + ... + 3⁴ + ... + 3²⁰²² + 3²⁰²³

3A - A = (3 + 3² + 3³ + ... + 3⁴ + 3²⁰²² + 3²⁰²³) - (1 + 3+...+ 3²⁰²²)

2A     = 3 + 3² + 3³ + 3⁴ + ... + 3²⁰²² + 3²⁰²³ - 1 - 3 - ... - 3²⁰²²

2A =  (3 - 3) + (3² - 3²) + (3⁴ - 3⁴) + (3²⁰²² - 3²⁰²²) + (3²⁰²³ - 1)

2A = 3²⁰²³ - 1 

 A  = \(\dfrac{3^{2023}-1}{2}\)

A = \(\dfrac{3^{2023}}{2}\) - \(\dfrac{1}{2}\)

B - A = \(\dfrac{3^{2023}}{2}\) - (\(\dfrac{3^{2023}}{2}\) - \(\dfrac{1}{2}\))

B - A = \(\dfrac{3^{2023}}{2}\) - \(\dfrac{3^{2023}}{2}\) + \(\dfrac{1}{2}\)

B - A = \(\dfrac{1}{2}\)

Lời giải:

$A=1+3+3^2+3^3+...+3^{2021}$

$3A=3+3^2+3^3+...+3^{2022}$

$\Rightarrow 3A-A=(3+3^2+3^3+...+3^{2022}) - (1+3+3^2+3^3+...+3^{2021})$

$\Rightarrow 2A=3^{2022}-1$

$\Rightarrow A=\frac{3^{2022}-1}{2}$

$B-A=\frac{3^{2022}}{2}-\frac{3^{2022}-1}{2}=\frac{1}{2}$

Đặt \(A=1+3+3^2+3^3+3^4+\cdot\cdot\cdot+3^{2023}+3^{2024}\)

\(=(1+3+3^2)+(3^3+3^4+3^5)+(3^6+3^7+3^8)+\dots+(3^{2022}+3^{2023}+3^{2024})\\=13+3^3\cdot(1+3+3^2)+3^6\cdot(1+3+3^2)+\dots+3^{2022}\cdot(1+3+3^2)\\=13+3^3\cdot13+3^6\cdot13+\dots+3^{2022}\cdot13\\=13\cdot(1+3^3+3^6+\dots+3^{2022})\)

Vì \(13\cdot(1+3^3+3^6+\dots+3^{2022})\vdots13\)

nên \(A\vdots13\)

\(\Rightarrowđpcm\)

Đặt S=1+3+3²+3³+3⁴+⋅⋅⋅+3²⁰²³+3²⁰²⁴

S=(1+3+3²)+(3³+3⁴+3⁵)+⋯+(3²⁰²²+3²⁰²³+3²⁰²⁴)

S=13+3³(1+3+3²)+...+3²⁰²²(1+3+3²)

S=13+3³.13+...+3²⁰²².13

S=13(3³+...+3²⁰²²) ⋮ 13

Vậy S⋮13

\(3^{2023}-3^{2021}=3^{2021}\left(3^2-1\right)=3^{2021}\cdot8⋮8\)