1. \(A=2^{2016}-1\)

\(2\equiv-1\left(mod3\right)\\ \Rightarrow2^{2016}\equiv1\left(mod3\right)\\ \Rightarrow2^{2016}-1\equiv0\left(mod3\right)\\ \Rightarrow A⋮3\)

\(2^{2016}=\left(2^4\right)^{504}=16^{504}\)

16 chia 5 dư 1 nên 16^504 chia 5 dư 1

=> 16^504-1 chia hết cho 5

hay A chia hết cho 5

\(2^{2016}-1=\left(2^3\right)^{672}-1=8^{672}-1⋮7\)

lý luận TT trg hợp A chia hết cho 5

(3;5;7)=1 = > A chia hết cho 105

2;3;4 TT ạ !!

A = 3¹ + 3² +3³ + 3⁴ +.....+3²⁰¹⁵+ 3²⁰¹⁶

A = (3¹ + 3²) +(3³ + 3⁴) +.....+ (3²⁰¹⁵+ 3²⁰¹⁶)

A = 3(1+3) + 3²(1+3) + .....+ 3²⁰¹⁵(1+3)

A = 3.4 +3².4 +....... + 3²⁰¹⁵.4

A = 4(3 +3² +....+ 3²⁰¹⁵) chia hết cho 4

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A =3¹ + 3² +3³ + 3⁴ + 3⁵ +3⁶ +.....+3²⁰¹⁴ + 3²⁰¹⁵+ 3²⁰¹⁶

A = (3¹ + 3² +3³ ) +(3⁴ + 3⁵ +3⁶) +.....+ (3²⁰¹⁴ + 3²⁰¹⁵+ 3²⁰¹⁶)

A = 3(1+3+3²) + 3⁴(1+3+3²) + .....+ 3²⁰¹⁴(1+3+3²)

A = 3.13 +3⁴.13 +....... + 3²⁰¹⁴.13

A = 13.(3 +3⁴ +....+ 3²⁰¹⁴) chia hết cho 13

\(A=3+3^2+...+3^{2016}\)

\(A=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2015}+3^{2016}\right)\)

\(A=3\cdot\left(1+3\right)+3^3\cdot\left(1+3\right)+...+3^{2015}\cdot\left(1+3\right)\)

\(A=4\cdot\left(3+3^3+...+3^{2015}\right)\)

Vậy A chia hết cho 4

_____________

\(A=3+3^2+3^3+...+3^{2016}\)

\(A=\left(3+3^2+3^3\right)+...+\left(3^{2014}+3^{2015}+3^{2016}\right)\)

\(A=3\cdot\left(1+3+9\right)+3^4\cdot\left(1+3+9\right)+...+3^{2014}\cdot\left(1+3+9\right)\)

\(A=13\cdot\left(3+3^4+...+3^{2014}\right)\)

Vậy A chia hết cho 13

Ta có \(A=4+2^2+2^3+...+2^{2016}\)

        \(A=2^2+2^2+2^3+...+2^{2016}\)

Ta có \(2^2+2^2=2^2.2=2^3\)

         \(2^3+2^3=2^3.2=2^4\) 

         ..........................................

Tương tự với các số hạng còn lại ta được 

     \(A=4+2^2+2^3+...+2^{2016}\)

    \(A=2^{2016}+2^{2016}=2^{2016}.2=2^{2017}\)chia hết cho \(2^{2017}\)

      Vậy A chia hết cho \(2^{2017}\)

Ta có 9911 = 11 . 17 . 53 . Trong mỗi tích đều có các thừa số đó :

- Tích các số lẻ có chứa các số 11 ; 17 ; 53

- Tích các số chẵn có các số 22 ; 34 ; 106 lần lượt là bội của các số 11 ; 17 ; 53

=> Tổng hai tích chia hết cho 9911.

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

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

             = 2 x 3 + 2³ x 3 + ... + 2⁵⁹ x 3.

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

Vì A = 3 x ( 2 + 2³ + ... + 2⁵⁹)  nên A chia hết cho 3.

           A= (2 +2² + 2³) + (2⁴ + 2⁵  + 2⁶) + ... + (2⁵⁸ + 2⁵⁹ + 2⁶⁰).

             = 2 x (1 + 2 + 2²) + 2⁴ x (1 + 2 + 2²) + ... + 2⁵⁸ x (1 + 2 + 2²).

             = 2 x 7 + 2⁴ x 7 + ... + 2⁵⁸ x 7.

             = 7 x ( 2 + 2⁴ + ... + 2⁵⁸).

Vì A = 7 x ( 2 + 2⁴ + ... + 2⁵⁸)  nên A chia hết cho 7.

  A= (2 +2² + 2³ + 2⁴) + (2⁵ + 2⁶  + 2⁷ + 2⁸) + ... + (2⁵⁷ + 2⁵⁸ + 2⁵⁹ + 2⁶⁰).

             = 2 x (1 + 2 + 2² + 2³) + 2⁵ x (1 + 2 + 2² + 2³) + ... + 2⁵⁷ x (1 + 2 + 2² + 2³).

             = 2 x 15 + 2⁵ x 15 + ... + 2⁵⁷ x 15.

             = 15 x ( 2 + 2⁴ + ... + 2⁵⁸).

Vì A = 15 x ( 2 + 2⁴ + ... + 2⁵⁸)  nên A chia hết cho 15.

Ta có: B= 3 + 3³ + 3⁵ + ... + 3¹⁹⁹¹= (3 + 3³ + 3⁵) + (3⁷+ 3⁹ + 3¹¹ ) + ... + (3¹⁹⁸⁷ + 3¹⁹⁸⁹ + 3¹⁹⁹¹).

             = 3 x (1 + 3² + 3⁴) + 3⁷ x (1 + 3² + 3⁴) + ... + 3¹⁹⁸⁷ x (1 + 3² + 3⁴).

             = 3 x 91 + 3⁷ x 91 + ... + 3¹⁹⁸⁷ x 91= 3 x 7 x 13 + 3⁷  x 7 x 13 + ... + 3¹⁹⁸⁷ x 7 x 13.

             = 13 x ( 3 x 7 + 3⁷ x 7 + ... + 3¹⁹⁸⁷ x 7).

Vì B = 13 x ( 3 x 7 + 3⁷ x 7 + ... + 3¹⁹⁸⁷ x 7) nên B chia hết cho 13.

           B= (3 + 3³ + 3⁵ + 3⁷) +  ... + (3¹⁹⁸⁵ + 3¹⁹⁸⁷ + 3¹⁹⁸⁹ + 3¹⁹⁹¹).

             = 3 x (1 + 3² + 3⁴  + 3⁶) +  ... + 3¹⁹⁸⁵ x (1 + 3² + 3⁴ ​+ 3⁶).

             = 3 x 820 + ... + 3¹⁹⁸⁵ x 820= 3 x 20 x 41 + ... + 3¹⁹⁸⁵ x 20 x 41.

             = 41 x ( 3 x 20 + .. +  3¹⁹⁸⁵ x 20)

Vì B =41 x ( 3 x 20 + .. +  3¹⁹⁸⁵ x 20) nên B chia hết cho 41.

a) Ta có: \(A=3+3^3+3^5+...+3^{1991}\)

\(=\left(3+3^3+3^5\right)+\left(3^7+3^9+3^{11}\right)+...+\left(3^{1987}+3^{1989}+3^{1991}\right)\)

\(=3\times\left(1+3^2+3^4\right)+3^7\times\left(1+3^2+3^4\right)+...+3^{1987}\times\left(1+3^2+3^4\right)\)

\(=3\times91+3^7\times91+...+3^{1987}\times91\)

\(=3\times7\times13+3^7\times7\times13+...+3^{1987}\times7\times13\)

\(=13\times\left(3\times7+3^7\times7+...+3^{1987}\times7\right)\)

Vì \(A=13\times\left(3\times7+3^7\times7+...+3^{1987}\times7\right)\)nên A chia hết cho 13.

b) Ta có: \(A=3+3^3+3^5+...+3^{1991}\)

\(=\left(3+3^3+3^5+3^7\right)+...+\left(3^{1985}+3^{1987}+3^{1989}+3^{1991}\right)\)

\(=3\times\left(1+3^2+3^4+3^6\right)+...+3^{1985}\times\left(1+3^2+3^4+3^6\right)\)

\(=3\times820+...+3^{1985}\times820\)

\(=3\times20\times41+...+3^{1985}\times20\times41\)

\(=41\times\left(3\times20+...+3^{1985}\times20\right)\)

Vì \(A=41\times\left(3\times20+...+3^{1985}\times20\right)\)nên A chia hết cho 41.

Đcm

Bài 1:Ta có:3¹⁵+3¹⁴=3¹⁴.3+3¹⁴=3¹⁴.4 chia hết cho 4

Bài 2:a,\(3A=3+3^2+3^3+...........+3^{2016}\)

\(\Rightarrow3A-A=\left(3+3^2+.......+3^{2016}\right)-\left(1+3+.......+3^{2015}\right)\)

\(\Rightarrow2A=3^{2016}-1\Rightarrow A=\frac{3^{2016}-1}{2}\)

b,Ta có:A=1+3+3²+3³+.............+3²⁰¹⁵

=(1+3)+(3²+3³)+...............+(3²⁰¹⁴+3²⁰¹⁵)

=4+3².4+................+3²⁰¹⁴.4

=4.(1+3²+.........+3²⁰¹⁴) chia hết cho 4

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