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:

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

đúng rùi

a) P = 1 + 3 + 3² + ... + 3¹⁰¹

= (1 + 3 + 3²) + (3³ + 3⁴ + 3⁵) + ... + (3⁹⁹ + 3¹⁰⁰ + 3¹⁰¹)

= 13 + 3³.(1 + 3 + 3²) + ... + 3⁹⁹.(1 + 3 + 3²)

= 13 + 3³.13 + ... + 3⁹⁹.13

= 13.(1 + 3³ + ... + 3⁹⁹) ⋮ 13

Vậy P ⋮ 13

b) B = 1 + 2² + 2⁴ + ... + 2²⁰²⁰

= (1 + 2² + 2⁴) + (2⁶ + 2⁸ + 2¹⁰) + ... + (2²⁰¹⁶ + 2²⁰¹⁸ + 2²⁰²⁰)

= 21 + 2⁶.(1 + 2² + 2⁴) + ... + 2²⁰¹⁶.(1 + 2² + 2⁴)

= 21 + 2⁶.21 + ... + 2²⁰¹⁶.21

= 21.(1 + 2⁶ + ... + 2²⁰¹⁶) ⋮ 21

Vậy B ⋮ 21

c) A = 2 + 2² + 2³ + ... + 2²⁰

= (2 + 2² + 2³ + 2⁴) + (2⁵ + 2⁶ + 2⁷ + 2⁸) + ... + (2¹⁷ + 2¹⁸ + 2¹⁹ + 2²⁰)

= 30 + 2⁴.(2 + 2² + 2³ + 2⁴) + ... + 2¹⁶.(2 + 2² + 2³ + 2⁴)

= 30 + 2⁴.30 + ... + 2¹⁶.30

= 30.(1 + 2⁴ + ... + 2¹⁶)

= 5.6.(1 + 2⁴ + ... + 2¹⁶) ⋮ 5

Vậy A ⋮ 5

d) A = 1 + 4 + 4² + ... + 4⁹⁸

= (1 + 4 + 4²) + (4³ + 4⁴ + 4⁵) + ... + (4⁹⁷ + 4⁹⁸ + 4⁹⁹)

= 21 + 4³.(1 + 4 + 4²) + ... + 4⁹⁷.(1 + 4 + 4²)

= 21 + 4³.21 + ... + 4⁹⁷.21

= 21.(1 + 4³ + ... + 4⁹⁷) ⋮ 21

Vậy A ⋮ 21

e) A = 11⁹ + 11⁸ + 11⁷ + ... + 11 + 1

= (11⁹ + 11⁸ + 11⁷ + 11⁶ + 11⁵) + (11⁴ + 11³ + 11² + 11 + 1)

= 11⁵.(11⁴ + 11³ + 11² + 11 + 1) + 16105

= 11⁵.16105 + 16105

= 16105.(11⁵ + 1)

= 5.3221.(11⁵ + 1) ⋮ 5

Vậy A ⋮ 5

A = 8⁸ + 2²⁰

= (2³)⁸ + 2²⁰

= 2²⁴ + 2²⁰

= 2²⁰.(2⁴ + 1)

= 2²⁰.17 ⋮ 17

Vậy A ⋮ 17

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 \(7\)

A= (2+22)+(23+24)+...+(259+260)
A=2.(1+2)+23.(1+2)+...+259.(1+2)
A=2.3+23.3+...+259.3
A=3.(2+23+...+259)
Vì 3 chia hết cho 3 => 3.(2+23+...+259)  chia hết cho 3
=>A  chia hết cho 3
A= (2+22+23)+...+(258+259+260)
A=2.(1+2+22)+...+258.(1+2+22)
A=2.7+...+258.7
A=7.(2+...+258)
Vì 7  chia hết cho 7 =>7.(2+...+258)  chia hết cho 7

CHIA HẾT CHO 3 :

A= (2+22)+(23+24)+...+(259+260)

A=2.(1+2)+23.(1+2)+...+259.(1+2)

A=2.3+23.3+...+259.3

A=3.(2+23+...+259)

Vì 3 chia hết cho 3 => 3.(2+23+...+259) chia hết cho 3

=>A chia hết cho 3

dcv

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

\(S=1+2+2^2+2^3+2^4+...+2^{2011}\)

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

\(\Rightarrow S=7+2^3.7+...+2^{2009}.7\)

\(\Rightarrow S=7\left(1+2^3+...+2^{2009}\right)⋮7\)

\(\Rightarrow dpcm\)

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:

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

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

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

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