\(B=2+2^2+2^3+...+2^{30}\)

\(=2\left(1+2\right)+2^3\left(1+2\right)+2^5\left(1+2\right)+...+2^{29}\left(1+2\right)\)

\(=2\cdot3+2^3\cdot3+2^5\cdot3+...+2^{29}\cdot3\)

\(=3\left(2+2^3+2^5+...+2^{29}\right)⋮3\)

Mặt khác:\(B=2+2^2+2^3+...+2^{30}\)

\(=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+....+2^{28}\left(1+2+2^2\right)\)

\(=2\cdot7+2^4\cdot7+....+2^{28}\cdot7\)

\(=7\left(2+2^4+...+2^{28}\right)⋮7\)

Mà (3;7)=1

\(\Rightarrow B⋮3\cdot7=21\)

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

c\m c\h cho 3:2+2^2+....+2^60=2.(1+2)+........+2^59(1+2)

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

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

                                              =>A chia hết cho 3

cau tiếp tuong tu

3

Ta chứng minh A chia hết cho 3:

A=(2+2^2)+(2^3+2^4)+...+(2^59+2^60)

  =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) chia hết cho 3

Ta chứng minh A chia hết cho 7

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

  =2.(1+2+4)+2^4.(1+2+4)+...+2^58.(1+2+4)

  =2.7+2^4.7+...+2^58.7

  =7.(2+2^4+...+2^58) chia hết cho 7

Ta chứng minh A chia hết cho 15

A=(2+2^2+2^3+2^4)+(2^5+2^6+2^7+2^8)+...+(2^57+2^58+2^59+2^60)

  =2.(1+2+4+8)+2^5.(1+2+4+8)+....+2^57.(1+2+4+8)

  =2.15+2^5.15+..+2^57.15

  =15.(2+2^5+...+2^57) chia hết cho 15

112 chia hết cho 7

nhưng : 2.1+3.1=5 không chia hết cho 7

B = 3¹ + 3² + 3³ + ... + 3²⁸ + 3²⁹ + 3³⁰

B = (  3¹ + 3² + 3³ ) + ... + ( 3²⁸ + 3²⁹ + 3³⁰ )

B = 3¹ . ( 1 + 3 + 3² ) + ... + 3²⁸ . ( 1 + 3 + 3² )

B = 3¹ . 13 + ... + 3²⁸ . 13

B = 13 . ( 3 + ... + 3²⁸ ) \(⋮\)13

Vậy B \(⋮\)13 ( dpcm )

\(B=3^1+3^2+3^3+3^4+3^5+............+3^{30}\)

\(\Rightarrow B=\left(3^1+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+............+\left(3^{28}+3^{29}+3^{30}\right)\)

\(\Rightarrow B=3^1.\left(1+3+3^2\right)+3^4.\left(1+3+3^2\right)+.........+3^{28}.\left(1+3+3^2\right)\)

\(\Rightarrow B=3^1.13+3^4.13+.........+3^{28}.13\)

\(\Rightarrow B=13\left(3^1+3^4+.........+3^{28}\right)\)

Mà 13 \(⋮\)13 \(\Rightarrow13\left(3^1+3^4+...........+3^{28}\right)⋮13\)

Vậy B chia hết cho 13

A la 7

B la 15

\(B=1+3+3^2+3^3+...+3^{98}+3^{99}\)

    \(=\left(1+3+3^2+3^3\right)+...+\left(3^{96}+3^{97}+3^{98}+3^{99}\right)\)

     \(=40+...+3^{96}.\left(1+3+3^2+3^3\right)\)

      \(=40+...+3^{96}.40\)

       \(=\left(1+...+3^{96}\right).40⋮40\)

\(\Rightarrow\) \(B⋮40\)

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