\(A=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{89}+3^{90}\right)\\ A=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{89}\left(1+3\right)\\ A=3\cdot4+3^3\cdot4+...+3^{89}\cdot4\\ A=4\left(3+3^3+...+3^{89}\right)⋮4\)

A = ( 3 + 3 2 ) + ( 3 3 + 3 4 ) + . . . + ( 3 89 + 3 90 )

A = 3 ( 1 + 3 ) + 3 3 ( 1 + 3 ) + . . . + 3 89 ( 1 + 3 )

A = 3 ⋅ 4 + 3 3 ⋅ 4 + . . . + 3 89 ⋅ 4

A = 4 ( 3 + 3 3 + . . . + 3 89 ) ⋮ 4

A = 8⁸ + 2²⁰

= (2³)⁸ + 2²⁰

= 2²⁴ + 2²⁰

= 2²⁰.(2⁴ + 1)

= 2²⁰.17 ⋮ 17

Vậy A ⋮ 17

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=3²+3³+3⁴+...+3⁹⁷
3A=3³+3⁴+3⁵+...+3⁹⁸

3A-A=(3³+3⁴+3⁵+...+3⁹⁸)-(3²+3³+3⁴+...+3⁹⁷)

2A=3⁹⁸-3²

A=(3⁹⁸-3²): 2

⇒A=(3⁹⁸-3²): 2

thế nhé chúc em học tốt :>>☺

ez

+) 3²+3³+3⁴+...+3⁹⁷

= (3²+3³)+...+ (3⁹⁶+3⁹⁷⁾

= 3².(1+3)+...+3⁹⁶.(1+3)

=3².4+...+3⁹⁶.4

=4.(3²+...+3⁹⁶)

Vì 4⋮4 nên 4.(3²+...+3⁹⁶)⋮4

+)P sau lm như p1 nhx là nhóm 3 số với nhau

còn chia hết cho 4 và 13 thì a wên cách làm r :>>

bạn hình như viết sai đề

`#3107.101107`

\(A=1+3+3^2+3^3+...+3^{101}\)

$A = (1 + 3 + 3^2) + (3^3 + 3^4 + 3^5) + ... + (3^{99} + 3^{100} + 3^{101}$

$A = (1 + 3 + 3^2) + 3^3 (1 + 3 + 3^2)  + ... + 3^{99}(1 + 3 + 3^2)$

$A = (1 + 3 + 3^2)(1 + 3^3 + ... + 3^{99})$

$A = 13(1 + 3^3 + ... + 3^{99})$

Vì `13(1 + 3^3 + ... + 3^{99}) \vdots 13`

`\Rightarrow A \vdots 13`

Vậy, `A \vdots 13.`

\(A=1+3+3^2+3^3+3^4+3^5+...+3^{101}\\=(1+3+3^2)+(3^3+3^4+3^5)+(3^6+3^7+3^8)+...+(3^{99}+3^{100}+3^{101})\\=13+3^3\cdot(1+3+3^2)+3^6\cdot(1+3+3^2)+...+3^{99}\cdot(1+3+3^2)\\=13+3^3\cdot13+3^6\cdot13+...+3^{99}\cdot13\\=13\cdot(1+3^3+3^6+...+3^{99})\)

Vì \(13\cdot(1+3^3+3^6...+3^{99}\vdots13\)

nên \(A\vdots13\)

\(\text{#}Toru\)

\(A=3+3^2+3^3+3^4+.......+3^{100}\)

\(\Rightarrow A=\left(3+3^2+3^3+3^4\right)+.......+\left(3^{97}+3^{98}+3^{99}+3^{100}\right)\)

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

\(\Rightarrow A=3.40+.........+3^{97}.40\)

\(\Rightarrow A=40.\left(3+.......+3^{97}\right)\)

\(\Rightarrow A⋮40\)( 1 )

Vì \(A\)là tổng của các bậc lũy thừa của 3 nên \(A⋮3\)( 2 )

Từ ( 1 ) và ( 2 ) suy ra : \(A⋮40.3\)

\(\Rightarrow A⋮120\)

Vậy \(A⋮120\)( ĐPCM )

mình ko biết

phải là chứng minh A chia hết cho 121

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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