a: \(A=1+2+2^2+...+2^{2023}\)

=>\(2A=2+2^2+2^3+...+2^{2024}\)

=>\(2A-A=2^{2024}+2^{2023}+...+2^2+2-2^{2023}-2^{2022}-...-2^2-2-1\)

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

b: \(A=\left(1+2\right)+2^2+2^3+...+2^{2023}\)

\(=3+2^2\left(1+2\right)+...+2^{2022}\left(1+2\right)\)

\(=3\left(1+2^2+...+2^{2022}\right)⋮3\)

Bài 1

a) S = 1 + 2 + 2² + 2³ + ... + 2²⁰²³

2S = 2 + 2² + 2³ + 2⁴ + ... + 2²⁰²⁴

S = 2S - S = (2 + 2² + 2³ + ... + 2²⁰²⁴) - (1 + 2 + 2² + 2³)

= 2²⁰²⁴ - 1

b) B = 2²⁰²⁴

B - 1 = 2²⁰²⁴ - 1 = S

B = S + 1

Vậy B > S

a,

\(S=1+2+2^2+...+2^{2023}\)

\(2S=2+2^2+2^3+...+2^{2024}\)

\(\Rightarrow S=2^{2024}-1\)

b.

Do \(2^{2024}-1< 2^{2024}\)

\(\Rightarrow S< B\)

2.

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

\(\Rightarrow3H=3^2+3^3+...+3^{2023}\)

\(\Rightarrow3H-H=3^{2023}-3\)

\(\Rightarrow2H=3^{2023}-3\)

\(\Rightarrow H=\dfrac{3^{2023}-3}{2}\)

Bài 2

H = 3 + 3² + 3³ + ... + 3²⁰²²

⇒ 3H = 3² + 3³ + 3⁴ + ... + 3²⁰²³

⇒2H = 3H - H

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

= 3²⁰²³ - 3

⇒ H = (3²⁰²³ - 3) : 2

2:

a: =>2(x+1)=26

=>x+1=13

=>x=12

b: =>(6x)^3=125

=>6x=5

=>x=5/6(loại)

c: =>\(7\cdot3^x\cdot\dfrac{1}{3}+11\cdot3^x\cdot3=318\)

=>3^x=9

=>x=2

d: -2x+13 chia hết cho x+1

=>-2x-2+15 chia hết cho x+1

=>15 chia hết cho x+1

=>x+1 thuộc {1;3;5;15}

=>x thuộc {0;2;4;14}

e: 4x+11 chia hết cho 3x+2

=>12x+33 chia hết cho 3x+2

=>12x+8+25 chia hết cho 3x+2

=>25 chia hết cho 3x+2

=>3x+2 thuộc {1;-1;5;-5;25;-25}

mà x là số tự nhiên

nên x=1

1: 

a: Đặt A=2^2024-2^2023-...-2^2-2-1

Đặt B=2^2023+2^2022+...+2^2+2+1

=>2B=2^2024+2^2023+...+2^3+2^2+2

=>B=2^2024-1

=>A=2^2024-2^2024+1=1

c: \(=\dfrac{3^{12}\cdot2^{11}+2^{10}\cdot3^{12}\cdot5}{2^2\cdot3\cdot3^{11}\cdot2^{11}}=\dfrac{2^{10}\cdot3^{12}\left(2+5\right)}{2^{13}\cdot3^{12}}\)

\(=\dfrac{7}{2^3}=\dfrac{7}{8}\)

Coi A=1+2+22+...+22024

B=5.22023
�=1+2+22+...+22022

$A=1+2+2^2+...+2^{2024}$

$\Rightarrow 2A=2+2^2+...+2^{2024}$

$\Rightarrow 2A-A=2^{2024}-1$

$\Rightarrow A=2^{2024}-1$

$\Rightarrow A<2^{2024}=2^{}{.2}^{2023}={2.2}^{2023}<{5.2}^{2023}$

$\Rightarrow A<B$

Ta có \(A=\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+...+\dfrac{2022}{2^{2022}}+\dfrac{2023}{2^{2023}}\)

\(2A=1+\dfrac{2}{2}+\dfrac{3}{2^2}+...+\dfrac{2022}{2^{2021}}+\dfrac{2023}{2^{2022}}\)

\(2A-A=\left(1+\dfrac{2}{2}+\dfrac{3}{2^2}+...+\dfrac{2022}{2^{2021}}+\dfrac{2023}{2^{2022}}\right)-\left(\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+...+\dfrac{2022}{2^{2022}}+\dfrac{2023}{2^{2023}}\right)\)\(A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2021}}+\dfrac{1}{2^{2022}}\) - \(\dfrac{2023}{2^{2023}}\)

Đặt B = \(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2021}}+\dfrac{1}{2^{2022}}\)

2B = \(2+1+\dfrac{1}{2}+...+\dfrac{1}{2^{2020}}+\dfrac{1}{2^{2021}}\)

2B - B = \(\left(2+1+\dfrac{1}{2}+...+\dfrac{1}{2^{2020}}+\dfrac{1}{2^{2021}}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2021}}+\dfrac{1}{2^{2022}}\right)\)B = 2 - \(\dfrac{1}{2^{2022}}\)

Suy ra  A = 2 - \(\dfrac{1}{2^{2022}}\) - \(\dfrac{2023}{2^{2023}}\) < 2

Vậy A < 2

\(A=\dfrac{1}{2}+\dfrac{2}{2^{2}}+\dfrac{3}{2^{3}}+...+\dfrac{2022}{2^{2022}}+\dfrac{2023}{2^{2023}}\)

\(2A=1+\dfrac22+\dfrac3{2^2}\ +\,.\!.\!.+\ \dfrac{2022}{2^{2021}}+\dfrac{2023}{2^{2022}}\\2A-A=\left(1+\dfrac22+\dfrac3{2^2}\ +\,.\!.\!.+\ \dfrac{2022}{2^{2021}}+\dfrac{2023}{2^{2022}}\right)-\left(\dfrac12+\dfrac2{2^2}+\dfrac3{2^3}\ +\,.\!.\!.+\ \dfrac{2022}{2^{2022}}+\dfrac{2023}{2^{2023}}\right)\\A=1+\dfrac12+\dfrac1{2^3}\ +\,.\!.\!.+\ \dfrac1{2^{2021}}+\dfrac1{2^{2022}}-\dfrac{2023}{2^{2023}}\\2\left(A+\dfrac{2023}{2^{2023}}\right)=2+1+\dfrac12+\dfrac1{2^2}\ +\,.\!.\!.+\ \dfrac1{2^{2020}}+\dfrac1{2^{2021}}\\A+\dfrac{2023}{2^{2023}}=2-\dfrac1{2^{2022}}\\A=2-\dfrac1{2^{2022}}+\dfrac{2023}{2^{2023}}<2\)

 

 

Sửa:

$2A=1+\dfrac22+\dfrac3{2^2}\ +\,.\!.\!.+\ \dfrac{2022}{2^{2021}}+\dfrac{2023}{2^{2022}}\\2A-A=\left(1+\dfrac22+\dfrac3{2^2}\ +\,.\!.\!.+\ \dfrac{2022}{2^{2021}}+\dfrac{20 23}{2^{2022}}\right)-\left(\dfrac12+\dfrac2{2^2}+\dfrac3{2^3}\ +\,.\!.\!.+\ \dfrac{2022}{2^{2022}}+\dfrac{2023}{2^{2023}}\right)\\A=1+\dfrac12+\dfrac1{2^3}\ +\,.\!.\!.+\ \dfrac1{2^{2021}}+\dfrac1{2^{2022}}-\dfrac{2023}{2^{2023}}\\2\left(A+\dfrac{2023}{2^{2023}}\right)=2+1+\dfrac12+\dfrac1{2^2}\ +\,.\!.\!.+\ \dfrac1{2^{2020}}+\dfrac1{2^{2021}}\\A+\dfrac{2023}{2^{2023}}=2-\dfrac1{2^{2022}}\\A=2-\dfrac1{2^{2022}}+\dfrac{2023}{2^{2023}}<2$

\(\Leftrightarrow2^2+2^3+2^4+...+2^n=2^{2003}-4\) (1)

Đặt vế trái là VT

\(2VT=2^3+2^4+2^5+...+2^{n+1}\)

\(\Rightarrow VT=2VT-VT=2^{n+1}-2^2\) Thay vào (1)

\(\Rightarrow\left(1\right)\Leftrightarrow2^{n+1}-4=2^{2023}-4\)

\(\Leftrightarrow2^{n+1}=2^{2023}\Rightarrow n+1=2023\Rightarrow n=2022\)

 

a) A = 2⁰ + 2¹ + 2² + 2³ + ... + 2²⁰²²

2A = 2 + 2² + 2³ + 2⁴ + ... + 2²⁰²³

A = 2A - A

= (2 + 2² + 2³ + 2⁴ + ... + 2²⁰²³) - (2⁰ + 2¹ + 2² + 2³ + ... + 2²⁰²²)

= 2²⁰²³ - 2⁰

= 2²⁰²³ - 1

Vậy A = B

b) A = 2021 . 2023

= (2022 - 1).(2022 + 1)

= 2022.(2022 + 1) - 2022 - 1

= 2022² + 2022 - 2022 - 1

= 2022² - 1 < 2022²

Vậy A < B

4A=2^2+2^4+...+2^2024

=>3A=2^2024-1

2B=2^2024

=>3A và 2B là hai số tự nhiên liên tiếp

a) P = 1 + 3 + 3² + ... + 3¹⁰¹

= (1 + 3 + 3²) + (3³ + 3⁴ + 3⁵) + ... + (3⁹⁹ + 3¹⁰⁰ + 3¹⁰¹)

= 13 + 3³.(1 + 3 + 3²) + ... + 3⁹⁹.(1 + 3 + 3²)

= 13 + 3³.13 + ... + 3⁹⁹.13

= 13.(1 + 3³ + ... + 3⁹⁹) ⋮ 13

Vậy P ⋮ 13

b) B = 1 + 2² + 2⁴ + ... + 2²⁰²⁰

= (1 + 2² + 2⁴) + (2⁶ + 2⁸ + 2¹⁰) + ... + (2²⁰¹⁶ + 2²⁰¹⁸ + 2²⁰²⁰)

= 21 + 2⁶.(1 + 2² + 2⁴) + ... + 2²⁰¹⁶.(1 + 2² + 2⁴)

= 21 + 2⁶.21 + ... + 2²⁰¹⁶.21

= 21.(1 + 2⁶ + ... + 2²⁰¹⁶) ⋮ 21

Vậy B ⋮ 21

c) A = 2 + 2² + 2³ + ... + 2²⁰

= (2 + 2² + 2³ + 2⁴) + (2⁵ + 2⁶ + 2⁷ + 2⁸) + ... + (2¹⁷ + 2¹⁸ + 2¹⁹ + 2²⁰)

= 30 + 2⁴.(2 + 2² + 2³ + 2⁴) + ... + 2¹⁶.(2 + 2² + 2³ + 2⁴)

= 30 + 2⁴.30 + ... + 2¹⁶.30

= 30.(1 + 2⁴ + ... + 2¹⁶)

= 5.6.(1 + 2⁴ + ... + 2¹⁶) ⋮ 5

Vậy A ⋮ 5

d) A = 1 + 4 + 4² + ... + 4⁹⁸

= (1 + 4 + 4²) + (4³ + 4⁴ + 4⁵) + ... + (4⁹⁷ + 4⁹⁸ + 4⁹⁹)

= 21 + 4³.(1 + 4 + 4²) + ... + 4⁹⁷.(1 + 4 + 4²)

= 21 + 4³.21 + ... + 4⁹⁷.21

= 21.(1 + 4³ + ... + 4⁹⁷) ⋮ 21

Vậy A ⋮ 21

e) A = 11⁹ + 11⁸ + 11⁷ + ... + 11 + 1

= (11⁹ + 11⁸ + 11⁷ + 11⁶ + 11⁵) + (11⁴ + 11³ + 11² + 11 + 1)

= 11⁵.(11⁴ + 11³ + 11² + 11 + 1) + 16105

= 11⁵.16105 + 16105

= 16105.(11⁵ + 1)

= 5.3221.(11⁵ + 1) ⋮ 5

Vậy A ⋮ 5