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)

`#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 \(A\vdots13\)

\(\text{#}Toru\)

\(S=\left(1+3+3^2\right)+...+3^7\left(1+3+3^2\right)\)

\(=13\left(1+...+3^7\right)⋮13\)

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 :>>

Đặt A = 3² + 3³ + 3⁴ + ... + 3⁹⁹

= 3² + 3³ + (3⁴ + 3⁵ + 3⁶) + (3⁷ + 3⁸ + 3⁹) + ... + (3⁹⁷ + 3⁹⁸ + 3⁹⁹)

= 36 + 3⁴.(1 + 3 + 3²) + 3⁷.(1 + 3 + 3²) + ... + 3⁹⁷.(1 + 3 + 3²)

= 36 + 3⁴.13 + 3⁷.13 + ... + 3⁹⁷.13

= 36 + 13.(3⁴ + 3⁷ + ... + 3⁹⁷)

Do 36 không chia hết cho 13

13.(3⁴ + 3⁷ + ... + 3⁹⁷) ⋮ 13

⇒ 36 + 13.(3⁴ + 3⁷ + ... + 3⁹⁷) không chia hết cho 13

⇒ A không chia hết cho 13

Em xem lại đề nhé, có thể em viết thiếu số 3 rồi

\(B=3+3^2+3^3+3^4+3^5+3^6+3^7+3^8\\=(3+3^2)+(3^3+3^4)+(3^5+3^6)+(3^7+3^8)\\=3\cdot(1+3)+3^3\cdot(1+3)+3^5\cdot(1+3)+3^7\cdot(1+3)\\=3\cdot4+3^3\cdot4+3^5\cdot4+3^7\cdot4\\=4\cdot(3+3^3+3^5+3^7)\)

Vì \(4\cdot(3+3^3+3^5+3^7) \vdots 4\)

nên \(B\vdots4\).

`#3107.101107`

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

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

\(=3\left(1+3\right)+3^3\left(1+3\right)+3^5\left(1+3\right)+3^7\left(1+3\right)\)

\(=\left(1+3\right)\left(3+3^3+3^5+3^7\right)\)

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

Vì \(4\left(3^3+3^5+3^7\right)\) $\vdots 4$

`\Rightarrow B \vdots 4`

Vậy, `B \vdots 4.`

B=3+32+33+34+35+36+37+38=(3+32)+(33+34)+(35+36)+(37+38)=3⋅(1+3)+33⋅(1+3)+35⋅(1+3)+37⋅(1+3)=3⋅4+33⋅4+35⋅4+37⋅4=4⋅(3+33+35+37)

Vì 4⋅(3+33+35+37)⋮44⋅(3+33+35+37)⋮4

nên $B⋮4$.

Ta có M = 3 + 3² + 3³ + 3⁴ + ... + 3¹⁸

              = ( 3 + 3² ) + ( 3³ + 3⁴ ) + ... + ( 3¹⁷ + 3¹⁸ )

              = 3( 1 + 3 ) + 3³( 1 + 3 ) + ... + 3¹⁷( 1 + 3 )

              = 3 . 4 + 3³ . 4 + ... + 3¹⁷ . 4

              = 4( 3 + 3³ + ... + 3¹⁷ ) ⋮ 4

Vậy M ⋮ 4

Lại có M = 3 + 3² + 3³ + 3⁴ + ... + 3¹⁸ 

               = ( 3 + 3² + 3³ ) + ( 3⁴ + 3⁵ + 3⁶ ) + ... + ( 3¹⁶ + 3¹⁷ + 3¹⁸ )

               = 3( 1 + 3 + 3² ) + 3⁴( 1 + 3 + 3² ) + ... + 3¹⁷( 1 + 3 + 3² ) 

               = 3 . 13 + 3⁴ . 13 + ... + 3¹⁷ . 13

               = 13( 3 + 3⁴ + ... + 3¹⁷ ) ⋮ 13

Vậy M ⋮ 4 và 13