A=(1+3+3²)+(3³+3⁴+3⁵)+...+(3⁵⁸+3⁵⁹+3⁶⁰)

=(1+3+9)+3³.(1+3+3²)+...+3⁵⁸.(1+3+3²)

=13+3³.13+...+3⁵⁸.13

=13.(1+3³+...+3⁵⁸) chia hết cho 13

=> A chia hết cho 13

=> đpcm.

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={2+2^2}+{2^3+2^4}+.......+{2^59+2^60}

={2.1+2.2}+{2^3.1+2^3.2}+....+{2^59.1+2^59.2}

=2{1+2}+2^3{1+2}+...+2^59{1+2}

=2.3+2^3.3+.....+2^59.3

=3.(2+2^3+...+2^59)

vi co thua so 3 => tich do chia het cho 3

A={2+2^2}+{2^3+2^4}+.......+{2^59+2^60}

={2.1+2.2}+{2^3.1+2^3.2}+....+{2^59.1+2^59.2}

=2{1+2}+2^3{1+2}+...+2^59{1+2}

=2.3+2^3.3+.....+2^59.3

=3.(2+2^3+...+2^59)

vi co thua so 3 => tich do chia het cho 3

Vì 13 là lẻ \(\Rightarrow\) 13, 13², 13³, 13⁴, 13⁵, 13⁶ là lẻ.

Mà lẻ + lẻ + lẻ + lẻ + lẻ + lẻ = chẵn nên 13 + 13² + 13³ + 13⁴ + 13⁵ + 13⁶ là chẵn. \(\Rightarrow\) 13 + 13² + 13³ + 13⁴ + 13⁵ + 13⁶ \(⋮\) 2

\(\Rightarrow\) ĐPCM

ta co :(3+3^2+3^3)+.......+(3^58+3^59+3^60)

=3(1+3+3^2)+.....+3^58(1+3+3^2)

=3.13+.........+3^58.13chia hết cho 13

A=3+3^2+3^3+....+3^60

=(3+3^2+3^3)+(3^4+3^5+3^6)+...+(3^58+3^59+3^60)

=(3+9+27)+(3^3.3+3^3.9+3^3.27)+...+(3^57.3+3^57.9+3^57.27)

=39+3^3.(3+9+27)+....+3^57.(3+9+27)

=39+3^3.39+....+3^57.39

vì 39:13 nên=>chia hết cho 13

a)Ta có: A=3+3²+3³+....+3⁶⁰

              = (3+3²)+(3³+3⁴)+.....+(3⁵⁹+3⁶⁰)

              = 3.(1+3)+3³.(1+3)+......+3⁵⁹.(1+3)

               = 4.(3+3³+....+3⁵⁹) chia hết cho 4

b) Làm tương tự. (Nhóm 3 số)

B=(3+3²+3³)+(3⁴+3⁵+3⁶)+...+(3⁵⁸+3⁵⁹+3⁶⁰)

=3(1+3+9)+3⁴(1+3+9)+...+3⁵⁸(1+3+9)

=13.3+13.3⁴+...+13.3⁵⁸

=13.(3+3⁴+...+3⁵⁸) luôn chia hết cho 13

vậy B chia hết cho 13

B=(3+3²)+(3³+3⁴)+...+(3⁵⁹+3⁶⁰)

B=3(1+3)+3³(1+3)+3⁴(1+3)+...+3⁵⁹(1+3)

4(4+3³+3⁴+...+3⁵⁹)

suy ra:4(4+3³+3⁴+...+3⁵⁹)chia hết cho 4

ta có :(3^1+3^2+3^3)+(3^4+3^5)+.......+(3^58+3^59+3^60)
=(3+9+27)+3^4.(3+9+27)+.......+3^58.(3+9+27)
=39+3^4.39+.......+3^58.39
=39.(3^1+3^4+..........+3^58) chia het cho 13
(Vi 39 chia het cho 13 )
Vậy A chia hết cho 13
thanks

b: \(A=3\left(1+3+3^2\right)+...+3^{58}\left(1+3+3^2\right)\)

\(=13\left(3+...+3^{58}\right)⋮13\)

\(a,\Leftrightarrow2A=8+2^3+2^4+...+2^{21}\\ \Leftrightarrow2A-A=8+2^3+2^4+...+2^{21}-4-2^2-2^3-...-2^{20}\\ \Leftrightarrow A=2^{21}+8-4-2^2=2^{21}\left(đpcm\right)\\ b,A=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{58}+3^{59}+3^{60}\right)\\ A=3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+...+3^{58}\left(1+3+3^2\right)\\ A=\left(1+3+3^2\right)\left(3+3^4+...+3^{58}\right)\\ A=13\left(3+3^4+...+3^{58}\right)⋮13\)

a,Chứng minh rằng A là một lũy thừa của 2

A=4+2^2+2^3+2^4+......+2^20

b,Chứng tỏ A=3^1+3^2+3^3+.....+3^60 chia hết cho 13

a ) 

B=(3+3²)+(3³+3⁴)+...+(3⁵⁹+3⁶⁰)

B=3(1+3)+3³(1+3)+3⁴(1+3)+...+3⁵⁹(1+3)

4(4+3³+3⁴+...+3⁵⁹)

suy ra:4(4+3³+3⁴+...+3⁵⁹)chia hết cho 4

b )

B=(3+3²+3³)+(3⁴+3⁵+3⁶)+...+(3⁵⁸+3⁵⁹+3⁶⁰)

=3(1+3+9)+3⁴(1+3+9)+...+3⁵⁸(1+3+9)

=13.3+13.3⁴+...+13.3⁵⁸

=13.(3+3⁴+...+3⁵⁸) luôn chia hết cho 13

vậy B chia hết cho 13

a) \(B=3+3^2+3^3+..+3^{60}\)

\(=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{59}+3^{60}\right)\)

\(=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{59}\left(1+3\right)\)

\(=4\left(3+3^3+...+3^{59}\right)⋮4\)

=>đpcm

b) \(B=3+3^2+3^3+..+3^{60}\)

\(=\left(3+3^2+3^3\right)+...+\left(3^{58}+3^{59}+3^{60}\right)\)

\(=3\left(1+3+3^2\right)+...+3^{58}\left(1+3+3^2\right)=13\left(3+..+3^{58}\right)⋮13\)

=>đpcm

a) \(B=3+3^2+3^3+...+3^{60}\)

\(B=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{59}+3^{60}\right)\)

\(B=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{59}\left(1+3\right)\)

\(B=3.4+3^3.4+...+3^{59}.4\)

\(B=\left(3+3^3+...+3^{59}\right).4⋮4\left(đpcm\right)\)

b) \(B=3+3^2+3^3+...+3^{60}\)

\(B=\left(3+3^2+3^3\right)+...+\left(3^{58}+3^{59}+3^{60}\right)\)

\(B=3\left(1+3+3^2\right)+...+3^{58}\left(1+3+3^2\right)\)

\(B=3.13+...+3^{58}.13\)

\(B=\left(3+...+3^{58}\right).13⋮13\left(đpcm\right)\)