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

=>3A=2^2024-1

2B=2^2024

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

Câu 1: 

$A=(2+2^2)+(2^3+2^4)+(2^5+2^6)+....+(2^{2019}+2^{2020})$

$=2(1+2)+2^3(1+2)+2^5(1+2)+....+2^{2019}(1+2)$

$=(1+2)(2+2^3+2^5+...+2^{2019})=3(2+2^3+2^5+...+2^{2019})\vdots 3$

-----------------

$A=2+(2^2+2^3+2^4)+(2^5+2^6+2^7)+....+(2^{2018}+2^{2019}+2^{2020})$

$=2+2^2(1+2+2^2)+2^5(1+2+2^2)+....+2^{2018}(1+2+2^2)$

$=2+(1+2+2^2)(2^2+2^5+....+2^{2018})$

$=2+7(2^2+2^5+...+2^{2018})$

$\Rightarrow A$ chia $7$ dư $2$.

Câu 2:

$B=(3+3^2)+(3^3+3^4)+....+(3^{2021}+3^{2022})$
$=3(1+3)+3^3(1+3)+...+3^{2021}(1+3)$

$=(1+3)(3+3^3+...+3^{2021})=4(3+3^3+....+3^{2021})\vdots 4$

-------------------

$B=(3+3^2+3^3)+(3^4+3^5+3^6)+...+(3^{2020}+3^{2021}+3^{2022})$

$=3(1+3+3^2)+3^4(1+3+3^2)+....+3^{2020}(1+3+3^2)$

$=(1+3+3^2)(3+3^4+...+3^{2020})=13(3+3^4+...+3^{2020})\vdots 13$ (đpcm)

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

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

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

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

\(=2+1=3\)

2²⁰²¹:2²⁰²⁰+2²⁰²²:2²⁰²⁰

         1         +        2

                   3

đây chứ đâu

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

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

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

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

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

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

\(\Rightarrow x-8=0\)

\(\Rightarrow x=8\)

Ta có: A = 4 + 2² + 2³ +2⁴ +.....+2²⁰¹⁹ +2²⁰²⁰

=> 2A = 8 +2³ +2⁴ + 2⁵ +.......+2²⁰²⁰ +2²⁰²¹

=> 2A - A = 2²⁰²¹ - 2³

=> A = 2²⁰²¹ - 2³

Yêu cầu của đề là gì thế bạn?