\(1,8^8+2^{20}=2^{24}+2^{20}=2^{20}\left(2^4+1\right)=2^{20}\cdot17⋮17\)

\(2,A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{119}+2^{120}\right)\\ A=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{119}\left(1+2\right)\\ A=3\left(2+2^3+...+2^{119}\right)⋮3\)

\(A=\left(2+2^2+2^3\right)+...+\left(2^{118}+2^{119}+2^{120}\right)\\ A=2\left(1+2+2^2\right)+...+2^{118}\left(1+2+2^2\right)\\ A=\left(1+2+2^2\right)\left(2+...+2^{118}\right)=7\left(2+...+2^{118}\right)⋮7\\ A=\left(2+2^2+2^3+2^4\right)+...+\left(2^{117}+2^{118}+2^{119}+2^{120}\right)\\ A=2\left(1+2+2^2+2^3\right)+...+2^{117}\left(1+2+2^2+2^3\right)\\ A=\left(1+2+2^2+2^3\right)\left(2+...+2^{117}\right)=15\left(2+...+2^{117}\right)⋮15\)

Mọi người giải giúp em với ạ. Em đang cần gấp !!!

1: \(A=2+2^2+2^3+2^4+...+2^{97}+2^{98}+2^{99}+2^{100}\)

\(=2\left(1+2+2^2+2^3\right)+...+2^{97}\left(1+2+2^2+2^3\right)\)

\(=15\left(2+2^5+...+2^{97}\right)\)

\(=30\left(1+2^4+...+2^{96}\right)⋮30\)

2:

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

\(=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2021}+3^{2022}\right)\)

\(=\left(3+3^2\right)+3^2\left(3+3^2\right)+...+3^{2020}\left(3+3^2\right)\)

\(=12\left(1+3^2+...+3^{2020}\right)⋮12\)

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 \(7\)

Sửa đề: \(A=2+2^2+...+2^{60}\)

\(=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+...+\left(2^{57}+2^{58}+2^{59}+2^{60}\right)\)

\(=2\left(1+2+2^2+2^3\right)+2^5\left(1+2+2^2+2^3\right)+...+2^{57}\left(1+2+2^2+2^3\right)\)

\(=15\left(2+2^5+...+2^{57}\right)\)

=>A chia hết cho 15 và A chia hết cho 3

\(A=2+2^2+...+2^{60}\)

\(=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{58}+2^{59}+2^{60}\right)\)

\(=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{58}\left(1+2+2^2\right)\)

\(=7\left(2+2^4+...+2^{58}\right)⋮7\)

Ta có: \(A⋮3;A⋮7\)

ƯCLN(3;7)=1

Do đó: \(A⋮3\cdot7=21\)

A = 8⁸ + 2²⁰

= (2³)⁸ + 2²⁰

= 2²⁴ + 2²⁰

= 2²⁰.(2⁴ + 1)

= 2²⁰.17 ⋮ 17

Vậy A ⋮ 17

giúp mình với mình chuẩn bị phải nộp bài rồi T~T 

\(B=2+2^2+2^3+...+2^{60}\)

\(=2\left(1+2+2^2\right)+...+2^{58}\left(1+2+2^2\right)\)

\(=7\cdot\left(2+...+2^{58}\right)⋮7\)

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)

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 3:

\(A=3+3^2+...+3^{100}\)

\(3A=3^2+3^3+...+3^{101}\)

\(3A-A=3^2+3^3+...+3^{101}-\left(3+3^2+...+3^{100}\right)\)

\(2A=3^{101}-3\) 

Mà: \(2A+3=3^N\)

\(\Rightarrow3^{101}-3+3=3^N\)

\(\Rightarrow3^{101}=3^N\)

\(\Rightarrow N=101\)

Vậy: ... 

Câu 1:

\(A=4+2^2+...+2^{20}\)

Đặt \(B=2^2+2^3+...+2^{20}\)

=>\(2B=2^3+2^4+...+2^{21}\)

=>\(2B-B=2^3+2^4+...+2^{21}-2^2-2^3-...-2^{20}\)

=>\(B=2^{21}-4\)

=>\(A=B+4=2^{21}-4+4=2^{21}\) là lũy thừa của 2

Câu 6:

Đặt A=1+2+3+...+n

Số số hạng là \(\dfrac{n-1}{1}+1=n-1+1=n\left(số\right)\)

=>\(A=\dfrac{n\left(n+1\right)}{2}\)

=>\(A⋮n+1\)

Câu 5:

\(A=5+5^2+...+5^8\)

\(=\left(5+5^2\right)+\left(5^3+5^4\right)+\left(5^5+5^6\right)+\left(5^7+5^8\right)\)

\(=\left(5+5^2\right)+5^2\left(5+5^2\right)+5^4\left(5+5^2\right)+5^6\left(5+5^2\right)\)

\(=30\left(1+5^2+5^4+5^6\right)⋮30\)