A =3+3²+3³+...+3¹¹⁹

A=(3+3²)+(3³+3⁴)+...(3¹¹⁸+3¹¹⁹)

A=3.(1+3)+3³.(1+3)+...+3¹¹⁸.(1+3)

A=3.4+3³.4+...+3¹¹⁸.4

A=4.(3+3³+...+3¹¹⁸)\(⋮\)4

=>A\(⋮\)4

A=3+3²+3³+...+3¹¹⁹

A=(3+3²+3³)+...+(3¹¹⁷+3¹¹⁸+3¹¹⁹)

A=3.(1+3+9)+...+3¹¹⁷.(1+3+9)

A=3.13+...+3¹¹⁷.13

A=13.(3+...+3¹¹⁷)\(⋮\)13

vì   A\(⋮\)4

và  A\(⋮\)13

=>A\(⋮\)4.13

=>A\(⋮\)52

vậy A\(⋮\)4 và A\(⋮\)52

A=1+3+3^2+3^3+...+3^98+3^99+3^100

A=(1+3+ 3^2)+(3^3+3^4+3^5)+...+(3^98+3^99+3^100)

A=(1+3+3^2)+3^3x(1+3+3^2)+...+3^98x(1+3+3^2)

A=13x3^3x13+...+3^98x13

=> 13x(1+3+3^3+...+3^98)chia hết cho 13

Vậy A chia hết cho 13

câu b đâu bạn ?

cho A = 1 + 3 + 3² + 3³ + ... +  3¹¹

^b) chứng minh A chia hết cho 40

A = 8⁸ + 2²⁰

= (2³)⁸ + 2²⁰

= 2²⁴ + 2²⁰

= 2²⁰.(2⁴ + 1)

= 2²⁰.17 ⋮ 17

Vậy A ⋮ 17

`#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,

A = 1 + 3 + 3² + 3³ + ... + 3¹¹⁹ 

3A = 3.(1 + 3 + 3² + 3³ + ... + 3¹¹⁹) 

3A = 3 + 3² + 3³ + 3⁴+ ... + 3¹²⁰

2A = 3A - A = (3 + 3² + 3³ + 3⁴ + ... + 3¹²⁰) - (1 + 3 + 3² + 3³ + ... + 3¹¹⁹) 

2A = 3¹²⁰ - 1 

A = \(\frac{3^{120}-1}{2}\)

Vậy A = \(\frac{3^{120}-1}{2}\)

b, Ta có : 3¹²⁰ - 1 + 1 = 27^x 

<=> 3¹²⁰ = 27^x 

<=> 3¹²⁰ = (3³)^x 

<=> 3¹²⁰ = 3^x 

<=> x = 120 

Vậy x = 120 

c, A có chia hết cho 5 và 13 

Sua cho \(\left(3^3\right)^x=3^{3x}\) nha 

\(\Rightarrow3^{120}=3^{3x}\Rightarrow x=\frac{120}{3}=40\)

C=(1+3+3^2)+3^3(1+3+3^2)+...+3^15(1+3+3^2)

=13(1+3^3+...+3^15) chia hết cho 13

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)

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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