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

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

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

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

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

A = 3 + 3² + 3³ + 3⁴ +..... + 3²⁰¹⁵ + 3²⁰¹⁶

= (3 + 3² + 3³) + (3⁴+ 3⁵ + 3⁶ ) +.....+  (3²⁰¹⁴ + 3²⁰¹⁵ + 3²⁰¹⁶)

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

= 13(3 + 3⁴ + ....+ 3²⁰¹⁴)  \(⋮13\)

A = 3 + 3² + 3³ + 3⁴ +..... + 3²⁰¹⁵ + 3²⁰¹⁶

= (3 + 3²) + (3³ + 3⁴) + .... + (3²⁰¹⁵ + 3²⁰¹⁶)

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

= 4(3 + 3³ + .... + 3²⁰¹⁵)     \(⋮4\)

A = 3 + 3² + 3³ + 3⁴ + ... + 3²⁰¹⁵ + 3²⁰¹⁶

A = (3 + 3²) + (3³ + 3⁴) + ... + (3²⁰¹⁵ + 3²⁰¹⁶)

A = 3(1 + 3) + 3³(1 + 3) + ... + 3²⁰¹⁵(1 + 3)

A = 3.4 + 3³.4 + ... + 3²⁰¹⁵.4

A = 4(3 + 3³ + ... + 3²⁰¹⁵)

Vì 4(3 + 3³ + ... + 3²⁰¹⁵) \(⋮\) 4 nên A \(⋮\) 4

Vậy A \(⋮\) 4

A = 3 + 3² + 3³ + 3⁴ + ... + 3²⁰¹⁵ + 3²⁰¹⁶

A = (3 + 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)

Vì 13(3 + 3⁴ + ... + 3²⁰¹⁴) \(⋮\) 13 nên A \(⋮\) 13

Vậy A \(⋮\) 13

*Chứng minh A chia hết cho 4

Ta có: \(A=\left(3^1+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2015}+3^{2016}\right)\)

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

\(=4\left(3^1+3^3+...+3^{2015}\right)⋮4^{\left(đpcm\right)}\)

*Chứng minh A chia hết cho 13

Ta có: \(A=\left(3^1+3^2+3^3\right)+...+\left(3^{2014}+3^{2015}+3^{2016}\right)\)

\(=3\left(1+3^1+3^2\right)+...+3^{2014}\left(1+3^1+3^2\right)\)

\(=13\left(3+...+3^{2014}\right)⋮13^{\left(đpcm\right)}\)

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 !!

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

\(A=1+3+3^2+3^3+3^4+...+3^{2015}\)

\(=\left(1+3+3^2+3^3\right)+\left(3^4+3^5+3^6+3^7\right)+...+\left(3^{2012}+3^{2013}+3^{2014}+3^{2015}\right)\)

\(=\left(1+3+3^2+3^3\right)+3^4\left(1+3+3^2+3^3\right)+...+3^{2012}\left(1+3+3^2+3^3\right)\)

\(=\left(1+3+3^2+3^3\right)\left(1+3^4+...+3^{2012}\right)\)

\(=40\left(1+3^4+...+3^{2012}\right)\)\(⋮\)\(5\)

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

\(=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+...+\left(2^{2013}+2^{2014}+2^{2015}+2^{2016}\right)\)

\(=2\left(1+2+2^2+2^3\right)+2^5\left(1+2+2^2+2^3\right)+..+2^{2013}\left(1+2+2^2+2^3\right)\)

\(=\left(1+2+2^2+2^3\right)\left(2+2^5+...+2^{2013}\right)\)

\(=15\left(2+2^5+...+2^{2013}\right)\)\(⋮\)\(15\)

giúp mình vs

alo

=(3+3^2+3^3+3^4)+(3^5+3^6+3^7+3^8)+.....+(3^2012+3^2013+3^2014+3^2015)

=3(1+3+9+27)+3^5(1+3+9+27)+.....+3^2012(1+3+9+27)

=40(3+3^5+...+3^2012)

=>A chia hết cho 10