B = 2+2²+2³+....+2³⁰

<=>  B=( 2 + 2²) + ( 2³ + 2⁴ )+....+(  2²⁹  +  2³⁰  ) 

<=> B=2. (1 + 2 ) + 2³. ( 1 + 2 )+ ...+ 2²⁹. ( 1  + 2 )

<=> B = 2.3 + 2³ + 3 +...+ 2²⁹.3

<=>B =3. ( 2 + 2³+... + 2²⁹)⋮ 3  (3)

Lại có B = 2+2²+2³+....+2³⁰

<=> B =( 2+2²+ 2³)+(2⁴+2⁵+2⁶ )+  ....+(2²⁸+2²⁹+2³⁰)

<=> B  =2.(1+2+2²)+2⁴ .(1+2+2²) + ..+2²⁸. (1+2+2²)

<=> B =2.7+ 2 ⁴.7 ..+ 2²⁸.7

<=> B =7.(2+...+)⋮ 7     (2) 

Mà (3;7) = 1   (3)

Từ (1); (2) và (3) <=> B ⋮ 21

Aiya tui lm sai đoạn cuối r

B = 2 + 2² + 2³ +. ...+ 2³⁰

<=> B =( 2 + 2²) + ( 2³ + 2⁴)+....+ ( 2²⁹ + 2³⁰)

<=> B = 2. (1 + 2  ) + 2³ .( 1 + 2 )+...+ 2²⁹. (1+2)

<=> B = 2. 3 + 2³ + 3 +...+ 2²⁹.3

<=> B = 3.(2 + 2³ + ... + 2²⁹)⋮ 3 (đpcm)    (1)

Mà B =  2 + 2² + 2³ + .... + 2³⁰

<=> B =( 2 + 2² + 2³) + (2⁴ + 2⁵ + 2⁶ )+....+ ( 2²⁸+2²⁹+2³⁰)

<=> B = 2. (1+2+2²)+ 2⁴.(1+2+2²) + ...+2²⁸.(1+2+2²)

<=> B = 2 .7 + 2⁴.7+...+2²⁸.7

<=> B =7.(2 + 2⁴+...+2²⁸ )⋮ 7 (đpcm)   (2) 

Lại có (3;7) = 1    (3)

Từ (1); (2) và (3) <=> B ⋮ 21

Nguồn mang + tớ nha 

Ta có: 21=3 x 7 vì 3 và 7 là 2 số nguyên tố cùng nhau

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

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

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

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

\(\Rightarrow B=3\left(2+2^3+...+2^{29}\right)\)

\(\Rightarrow B⋮3\left(1\right)\)

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

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

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

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

\(\Rightarrow B=7\left(2+2^4+....+2^{28}\right)\)

\(\Rightarrow B⋮7\left(2\right)\)

(1) (2) => B chia hết cho 21 (đpcm)

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

\(B=\left(2+2^2+2^3+2^4+2^5+2^6\right)+....+\left(2^{25}+2^{26}+2^{27}+2^{28}+2^{29}+2^{30}\right)\)

\(B=\left(2+2^2+2^3+2^4+2^5+2^6\right)+....+25^{24}\left(2+2^2+2^3+2^4+2^5+2^6\right)\)

\(B=126+...+2^{24}.126\)

\(B=126\left(1+...+2^{24}\right)\)

\(\Rightarrow B=126\left(1+...+2^{24}\right)⋮21\)

Vậy \(B⋮21\)

Hok tốt!!!

Nếu B chia hết cho 21 suy ra B chia hết cho 3,7

B=( 2+2^2)+(2^3+2^4)+......+( 2^9+2^30)

  = 2(1+2)+2^3(1+2)+.......+2^9(1+2)

  =2.3+2^3.3+....+2^29.3

  =3(2+2^3+........+2^29) chia hét cho 3

B=(2+2^2+2^3)+.....+(2^28+2^29+2^30)

  =2(1+2+2^2)+........+2^28(1+2+2^2)

  =2.7+.....+2^28.7

  =7(2+......+2^28) chia hết cho 7

Vậy B chia hết cho 7

a: \(A=2\left(1+2+2^2\right)+...+2^{19}\left(1+2+2^2\right)\)

\(=7\left(2+...+2^{19}\right)⋮7\)

tự làm nha