A=4+2²+2³+....+2²⁰

2A=8+2³+2⁴+...+2²¹

=> A+2A-A = (8+2³+2⁴+...+2²¹)  - (4+2²+2³+....+2²⁰)

=>A=2²¹+8 - (2²+4)=2²¹

=>A là 1 lũy thừa của 2

A= 4+22+23+....+220

2A= 8+23+24+...+221

A + 2A  -A = (8+2^3+2^4+...+2^21)  - (4+2^2+2^3+....+2^20)

A= 2^21+8 - (2^2+4)=2^21

Vậy A là 1 lũy thừa của 2

Ta có : 

\(A=2+2^2+2^3+2^4...2^{2010}\)\(^0\)

\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{2009}\left(1+2\right)\)

\(=2.3+2^3.3+....+2^{2009}.3\)

\(=3\left(2+2^3+....+2^{2009}\right)⋮3\)

Ta có :

\(2+2^2+2^3+2^4+....+2^{2010}\)

\(=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{2008}\left(1+2+2^2\right)\)

\(=2.7+2^4.7+....+2^{2008}.7\)

\(=7\left(2+2^4+....+2^{2008}\right)⋮7\)

Vậy \(2^1+2^2+2^3+2^4+...+2^{2010}⋮3\) và \(7\)

ĐỀ ĐÚNG KHÔNG ĐẤY SAO LẠI CÓ 23 

Mình ko biết, thầy ra mà

vậy nên mình mới hỏi chứ bỏ 23 đi thì mình tự giải cũng đc

TK :

A=(2+2²)+(2³+2⁴)+...+(2²⁰⁰⁹+2²⁰¹⁰)

A=(1+2)(2+2³+...+2²⁰⁰⁹)=3(2+...+2²⁰⁰⁹)⋮3

A=(2+2²+2³)+...+(2²⁰⁰⁸+2²⁰⁰⁹+2²⁰¹⁰ )

A=(1+2+2²)(2+...+2²⁰⁰⁸)=7(2+...+2²⁰⁰⁸)⋮7

Em xem lại đề nhé vì A như thế không chia hết cho 3 và cho 7

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=(1+2+2^2)+2^3(1+2+2^2)+...+2^96(1+2+2^2)+2^99

=7(1+2^3+...+2^96)+2^99 ko chia hết cho 7

Đề sai, viết lại thành:

A= 2¹+2²+2³+2⁴+...+2⁵⁹+2⁶⁰

Giải:

A=2¹+2²+2³+...............+2⁵⁹+2⁶⁰

A=(2¹+2²+2³)+...............+(2⁵⁸+2⁵⁹+2⁶⁰)

A=2.(1+2+2²)+............+2⁵⁸.(1+2+2²)

A=2.7+.......................+2⁵⁸.7

A=(2+2⁴+..............+2⁵⁸).7 ⋮ 7(đpcm)

Bài 1:

\(a,A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{2009}+2^{2010}\right)\\ A=\left(1+2\right)\left(2+2^3+...+2^{2009}\right)=3\left(2+...+2^{2009}\right)⋮3\\ A=\left(2+2^2+2^3\right)+...+\left(2^{2008}+2^{2009}+2^{2010}\right)\\ A=\left(1+2+2^2\right)\left(2+...+2^{2008}\right)=7\left(2+...+2^{2008}\right)⋮7\)

\(b,\left(\text{sửa lại đề}\right)B=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2009}+3^{2010}\right)\\ B=\left(1+3\right)\left(3+3^3+...+3^{2009}\right)=4\left(3+3^3+...+3^{2009}\right)⋮4\\ B=\left(3+3^2+3^3\right)+...+\left(3^{2008}+3^{2009}+3^{2010}\right)\\ B=\left(1+3+3^2\right)\left(3+...+3^{2008}\right)=13\left(3+...+3^{2008}\right)⋮13\)

Bài 2:

\(a,\Rightarrow2A=2+2^2+...+2^{2012}\\ \Rightarrow2A-A=2+2^2+...+2^{2012}-1-2-2^2-...-2^{2011}\\ \Rightarrow A=2^{2012}-1>2^{2011}-1=B\\ b,A=\left(2020-1\right)\left(2020+1\right)=2020^2-2020+2020-1=2020^2-1< B\)

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