\(A=2+2^2+2^3+...+2^{2020}+2^{2021}+2^{2022}\\=(2+2^2)+(2^3+2^4)+(2^5+2^6)+...+(2^{2021}+2^{2022})\\=2\cdot(1+2)+2^3\cdot(1+2)+2^5\cdot(1+2)+...+2^{2021}\cdot(1+2)\\=2\cdot3+2^3\cdot3+2^5\cdot3+...+2^{2021}\cdot3\\=3\cdot(2+2^3+2^5+..+2^{2021})\)

Vì \(3\cdot\left(2+2^3+2^5+...+2^{2021}\right)⋮3\)

nên \(A⋮3\).

\(Toru\)

A=(2+2²)+2²(2+2²)+...+2²⁰²⁰(2+2²)

A= 6.1+2².6+...+2²⁰²⁰.6

A=6(1+2²+...+2²⁰²⁰) chia hết cho 3

vậy A chia hết cho 3

A=(2+2²)+(2³+2⁴)+(2⁵+2⁶)+.......+(2²⁰¹⁹+2²⁰²⁰)+(2²⁰²¹+2²⁰²²)

A=2.(1+2)+2³.(1+2)+2⁵.(1+2)+.......+2²⁰¹⁹.(1+2)+2²⁰²¹.(1+2)

A=2.3+2³.3+2⁵.3+.......+2²⁰¹⁹.3+2²⁰²¹.3

A=3.(2+2³+2⁵+........+2²⁰¹⁹+2²⁰²¹)

Vì 3⋮3⇒A⋮3

a) Đặt A =  2.11 + 2.13 + ... + 2.29

= 2.(11 + 13 + 15 + ... + 29)

Đặt B = 11 + 13 + 15 + ... + 29

Số số hạng của B:

(29 - 11) : 2 + 1 = 10 (số)

A = 2.(29 + 11) . 10 : 2

= 40.10

= 400

b) (2²⁰²² + 2²⁰²¹- 2²⁰²⁰) : (2²⁰¹⁹ . 2)

= 2²⁰²⁰.(2² + 2 - 1) : 2²⁰²⁰

= 4 + 2 - 1

= 5

\(A=1+2+2^2+...+2^{2020}+2^{2021}\\ \Rightarrow2A=2+2^2+2^3+...+2^{2021}+2^{2022}\\ \Rightarrow2A-A=A=2^{2022}-1\)

Vậy \(A\) và \(B\) là 2 số tự nhiên liên tiếp.

\(\left(x-4\right)\cdot2^{2020}=2^{2022}\)

\(\Rightarrow\left(x-4\right)\cdot2^{2020}-2^{2022}=0\)

\(\Rightarrow2^{2020}\cdot\left[\left(x-4\right)-2^2\right]=0\)

\(\Rightarrow2^{2020}\cdot\left(x-4-4\right)=0\)

\(\Rightarrow2^{2020}\cdot\left(x-8\right)=0\)

\(\Rightarrow x-8=0\)

\(\Rightarrow x=8\)

A = 1 + 2 + 2² + ... + 2²⁰²¹
2A = 2 + 4 + 2³ + ... 2²⁰²²
A = 2²⁰²² - 1

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

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

\(2A-A=\left(2+2^2+2^3+...+2^{2021}+2^{2022}\right)-\left(1+2+2^2+...+2^{2020}+2^{2021}\right)\)

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

2A=2*(1+2+2²+...+2²⁰²⁰)=2+2²+...+2²⁰²¹

^2A-A=(1+2+2²+...+2²⁰²¹)-(1+2+2²+...+2²⁰²⁰)

A=2²⁰²¹-1<2021

Giải:

A=1+2+2²+2³+...+2²⁰²⁰

2A=2+2²+2³+2⁴+...+2²⁰²¹

2A-A=(2+2²+2³+2⁴+...+2²⁰²¹)-(1+2+2²+2³+...+2²⁰²⁰)

A=2²⁰²¹-1

⇒A<2²⁰²¹

Chúc bạn học tốt!

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

\(\Rightarrow\dfrac{1}{2}A=\dfrac{1}{2}.\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2020}}+\dfrac{1}{2^{2021}}\right)\)\(\Rightarrow\dfrac{1}{2}A=\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{2021}}+\dfrac{1}{2^{2022}}\)

\(\Rightarrow A-\dfrac{1}{2}A=\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2020}}+\dfrac{1}{2^{2021}}\right)-\left(\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{2021}}+\dfrac{1}{2^{2022}}\right)\)\(\Rightarrow\dfrac{1}{2}A=\dfrac{1}{2}-\dfrac{1}{2^{2022}}\)

\(\Rightarrow\dfrac{1}{2}A=\dfrac{2^{2021}-1}{2^{2022}}\)

\(\Rightarrow A=\dfrac{2^{2021}-1}{2^{2023}}.2=\dfrac{2^{2021}-1}{2^{2021}}\)

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