\(A=3+3^2+3^3+3^4+...+3^{90}\)

\(=\left(3+3^2+3^3+3^4+3^5\right)+...+\left(3^{86}+3^{87}+3^{88}+3^{89}+3^{90}\right)\)

\(=3.\left(1+3+3^2+3^3+3^4\right)+...+3^{86}\left(1+3+3^2+3^3+3^4\right)\)

\(=3.121+...+3^{36}.121\)

\(=121\left(3+...+3^{86}\right)⋮11\left(dpcm\right)\)

\(A=3+3^2+3^3+3^4+...+3^{90}\)

\(=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{88}+3^{89}+3^{90}\right)\)

\(=\left(3+3^2+3^3\right)+\left(3^3.3+3^3.3^2+3^3.3^3\right)+...+\left(3^{87}.3+3^{87}.3^2+3^{87}.3\right)\)

\(=\left(3+3^2+3^3\right)+3^3\left(3+3^2+3^3\right)+...+3^{87}\left(3+3^2+3^3\right)\)

\(=39.1+3^3.39+...3^{87}.39\)

\(=39\left(3^3+1+...+3^{87}\right)\)

\(=13.3\left(3^3+1+...+3^{87}\right)⋮13\left(dpcm\right)\)

TA CÓ:

A=3⁰+3+3²+3³+........+3¹¹

(3⁰+3+3²+3³)+....+(3⁸+3⁹+3¹⁰+3¹¹)

3(0+1+3+3²)+......+3⁸(0+1+3+3²) 

3.13+....+3⁸.13 cHIA HẾT CHO 13 NÊN A CHIA HẾT CHO 13( đpcm)

Lời giải:

$A=(3+3^2+3^3)+(3^4+3^5+3^6)+....+(3^{88}+3^{89}+3^{90})$

$=3(1+3+3^2)+3^4(1+3+3^2)+...+3^{88}(1+3+3^2)$

$=(1+3+3^2)(3+3^4+...+3^{88})=13(3+3^4+...+3^{88})\vdots 13$
--------------------

$A=(3+3^2+3^3+3^4+3^5)+(3^6+3^7+3^8+3^9+3^{10})+...+(3^{86}+3^{87}+3^{88}+3^{89}+3^{90})$
$=3(1+3+3^2+3^3+3^4)+3^6(1+3+3^2+3^3+3^4)+...+3^{86}(1+3+3^2+3^3+3^4)$
$=(1+3+3^2+3^3+3^4)(3+3^6+...+3^{86})$

$=121(3+3^6+...+3^{86})=11.11.(3+3^6+...+3^{86})\vdots 11$

\(A=1+3+3^2+..........+3^{11}\)

\(\Leftrightarrow A=\left(1+3\right)+\left(3^2+3^3\right)+.........+\left(3^{10}+3^{11}\right)\)

\(\Leftrightarrow A=1\left(1+3\right)+3^2\left(1+3\right)+.........+3^{10}\left(1+3\right)\)

\(\Leftrightarrow A=1.4+3^2.4+.......+3^{10}.4\)

\(\Leftrightarrow A=4\left(1+3^2+..........+3^{10}\right)⋮4\left(đpcm\right)\)

A = 1 + 3 + 3² + 3³ + ... + 3¹¹

A = ( 1 + 3 ) + ( 3² + 3³ ) + ... + ( 3¹⁰ + 3¹¹ )

A = 4 + 3² . ( 1 + 3 ) + ... + 3¹⁰ . ( 1 + 3 )

A = 4 + 3² . 4 + ... + 3¹⁰ . 4

A = 4 . ( 1 + 3² + ... + 3¹⁰ ) \(⋮\) 4 ( Vì trong tích có một thừa số chia hết cho 4 )

~ Chúc bạn học giỏi ! ~

B = (1 + 3) + (3²+3³)+.....+(3⁸⁹+3⁹⁰)

  = 4 + 3² .(1 + 3) + .....+3⁹⁰.(1+3)

 = 1 .4 + 3².4 + ..... +3⁹⁰.4

= 4.(1 + 3² + .... +3⁹⁰) chia hết cho 4

\(S=3+3^2+3^3+3^4+....+3^{89}+3^{90}\)

\(=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{88}+3^{89}+3^{90}\right)\)

\(==3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+3^{88}\left(1+3+3^2\right)\)

\(=\left(1+3+3^2\right).\left(3+3^4+....+3^{88}\right)\)

\(=13\left(3+3^4+...+3^{88}\right)\)\(⋮\)\(13\)

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.

a) Ta có : 

A = 1 + 3 + 3² + .... + 3¹¹

A = (1 + 3 + 3²) + (3³ + 3⁴ + 3⁵) + (3⁶ + 3⁷ + 3⁸) + (3⁹ + 3¹⁰ + 3¹¹)

A = 1 . (1 + 3 + 9) + 3³ . (1 + 3 + 9) + 3⁶ . (1 + 3 + 9) + 3⁹ . (1 + 3 + 9)

A = 1. 13 + 3³ . 13 + 3⁶ . 13 + 3⁹ . 13

A = 13 . (1 + 3³ + 3⁶ + 3⁹) chia hết cho 13 (ĐPCM)

b) Ta có : 

A = 1 + 3 + 3² + 3³ + ... + 3¹¹

A = (1 + 3 + 3² + 3³) + (3⁴ + 3⁵ + 3⁶ + 3⁷) + (3⁸ + 3⁹ + 3¹⁰ + 3¹¹)

A = 1 . (1 + 3 + 9 + 27) + 3⁴ . (1 + 3 + 9 + 27) + 3⁸ . (1 + 3 + 9 + 27)

A = 1 . 40 + 3⁴ . 40 + 3⁸ . 40

A = 40 . (1 + 3⁴ + 3⁸) chia hết cho 40 (ĐPCM)

Ủng hộ mk nha !!! ^_^

a) Ta có : 

A = 1 + 3 + 3² + .... + 3¹¹

A = (1 + 3 + 3²) + (3³ + 3⁴ + 3⁵) + (3⁶ + 3⁷ + 3⁸) + (3⁹ + 3¹⁰ + 3¹¹)

A = 1 . (1 + 3 + 9) + 3³ . (1 + 3 + 9) + 3⁶ . (1 + 3 + 9) + 3⁹ . (1 + 3 + 9)

A = 1. 13 + 3³ . 13 + 3⁶ . 13 + 3⁹ . 13

A = 13 . (1 + 3³ + 3⁶ + 3⁹) chia hết cho 13 (ĐPCM)

b) Ta có : 

A = 1 + 3 + 3² + 3³ + ... + 3¹¹

A = (1 + 3 + 3² + 3³) + (3⁴ + 3⁵ + 3⁶ + 3⁷) + (3⁸ + 3⁹ + 3¹⁰ + 3¹¹)

A = 1 . (1 + 3 + 9 + 27) + 3⁴ . (1 + 3 + 9 + 27) + 3⁸ . (1 + 3 + 9 + 27)

A = 1 . 40 + 3⁴ . 40 + 3⁸ . 40

A = 40 . (1 + 3⁴ + 3⁸) chia hết cho 40 (ĐPCM)

a) \(B=3+3^2+...+3^{90}\)

\(\Leftrightarrow B=\left(3+3^2\right)+...+\left(3^{89}+3^{90}\right)\)

\(\Leftrightarrow B=\left(3+3^2\right)+...+3^{88}.\left(3+3^2\right)\)

\(\Leftrightarrow B=12+...+3^{88}.12\)

\(\Leftrightarrow B=12.\left(1+...+3^{88}\right)⋮4\left(đpcm\right)\)

b)\(B=3+3^2+...+3^{90}\)

\(\Leftrightarrow B=\left(3+3^2\right)+...+\left(3^{89}+3^{90}\right)\)

\(\Leftrightarrow B=\left(3+3^2\right)+...+3^{88}.\left(3+3^2\right)\)

\(\Leftrightarrow B=12+...+3^{88}.12\)

\(\Leftrightarrow B=12.\left(1+...+3^{88}\right)⋮12\left(đpcm\right)\)

c) \(B=3+3^2+...+3^{90}\)

\(\Leftrightarrow B=\left(3+3^2+3^3\right)+...+\left(3^{88}+3^{89}+3^{90}\right)\)

\(\Leftrightarrow B=\left(3+3^2+3^3\right)+...+3^{87}.\left(3+3^2+3^3\right)\)

\(\Leftrightarrow B=39+...+3^{87}.39\)

\(\Leftrightarrow B=39.\left(1+..+3^{87}\right)⋮39\left(đpcm\right)\)

(1+3+3^2)+(3^3+3^4+3^5)+...+(3^9+3^10+3^11)

13+13.3+13.3^2+...+13.3^4

VẬY A CHIA HẾT CHO 13

Thank's