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

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

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

\(A=4+3^2\cdot4+....+3^{10}\cdot4\)

\(A=4\cdot\left(1+3^2+...+3^{10}\right)\) chia hết cho 4

Vì ta có 4 chia hết cho 4 => A có chia hết cho 4

Vậy A chia hết cho 4

2:

\(C=5+5^2+5^3+...+5^8\) chia hết cho 30

\(C=\left(5+5^2\right)+...+\left(5^7+5^8\right)\)

\(C=30+5^2\cdot\left(5+5^2\right)+...+5^6\cdot\left(5+5^2\right)\)

\(C=30\cdot1+5^2\cdot30+...5^6\cdot30\)

\(C=30\cdot\left(5^2+...+5^6\right)\)

Vì ta có 30 chia hết cho 30 nên suy ra C có chia hết cho 30

Vậy C có chia hết cho 30

a) \(C=5+5^2+5^3+...+5^8\)

\(C=\left(5+5^2\right)+\left(5^3+5^4\right)+\left(5^5+5^6\right)+\left(5^7+5^8\right)\)

\(C=\left(5+25\right)+5^2\cdot\left(5+25\right)+5^4\cdot\left(5+25\right)+5^6\cdot\left(5+25\right)\)

\(C=30+5^2\cdot30+5^4\cdot30+5^6\cdot30\)

\(C=30\cdot\left(1+5^2+5^4+5^6\right)\)

Vậy C chia hết cho 30

b) \(D=2+2^2+2^3+...+2^{60}\)

\(D=2\left(1+2\right)+2^2\left(1+2\right)+...+2^{59}\cdot\left(1+2\right)\)

\(D=2\cdot3+2^2\cdot3+...+2^{59}\cdot3\)

\(D=3\cdot\left(2+2^2+...+2^{59}\right)\)

Vậy D chia hết cho 3

\(D=2+2^2+2^3+...+2^{60}\)

\(D=2\cdot\left(1+2+4\right)+2^4\cdot\left(1+2+4\right)+...+2^{58}\cdot\left(1+2+4\right)\)

\(D=2\cdot7+2^4\cdot7+...+2^{58}\cdot7\)

\(D=7\cdot\left(2+2^4+...+2^{58}\right)\)

Vậy D chia hết cho 7

\(D=2+2^2+2^3+...+2^{60}\)

\(D=\left(2+2^2+2^3+2^4\right)+....+\left(2^{57}+2^{58}+2^{59}+2^{60}\right)\)

\(D=2\cdot\left(1+2+4+8\right)+...+2^{57}\cdot\left(1+2+4+8\right)\)

\(D=2\cdot15+2^5\cdot15+...+2^{57}\cdot15\)

\(D=15\cdot\left(2+2^5+...+2^{57}\right)\)

Vậy D chia hết cho 15 

a) C = 5 + 5² + 5³ + ... + 5⁸

= (5 + 5²) + 5².(5 + 5²) + 5⁴.(5 + 5²) + 5⁶.(5 + 5²)

= 30 + 5².30 + 5⁴.30 + 5⁶.30

= 30.(1 + 5² + 5⁴ + 5⁶) ⋮ 30

Vậy C ⋮ 30

b) *) Chứng minh D ⋮ 3

D = 2 + 2² + 2³ + ... + 2⁶⁰

= 2.(1 + 2) + 2³.(1 + 2) + ... + 2⁵⁹.(1 + 2)

= 2.3 + 2³.3 + ... + 2⁵⁹.3

= 3.(2 + 2³ + ... + 2⁵⁹) ⋮ 3

Vậy D ⋮ 3   (1)

*) Chứng minh D ⋮ 7

D = 2 + 2² + 2³ + ... + 2⁶⁰

= 2.(1 + 2 + 2²) + 2⁴.(1 + 2 + 2²) + ... 2⁵⁸.(1 + 2 + 2²)

= 2.7 + 2⁴.7 + ... + 2⁵⁸.7

= 7.(2 + 2⁴ + ... + 2⁵⁸) ⋮ 7

Vậy D ⋮ 7   (2)

*) Chứng minh D ⋮ 15

D = 2 + 2² + 2³ + ... + 2⁶⁰

= 2.(1 + 2 + 2² + 2³) + 2⁵.(1 + 2 + 2² + 2³) + 2⁵⁷.(1 + 2 + 2² + 2³)

= 2.15 + 2⁵.15 + ... + 2⁵⁷.15

= 15.(2 + 2⁵ + ... + 2⁵⁷) ⋮ 15

Vậy D ⋮ 15   (3)

Từ (1), (2), (3) suy ra D chia hết cho lần lượt 3; 7 và 15

A = 5 + 5^2 + 5^3 +...+5^8 

A = ( 5 + 5^2 ) + ( 5^3 + 5^4) +...+(5^7+5^8)

A = 30 +5(5+5^2) +...+ 5^6(5 + 5^2)

A=30 + 5 . 30 +...+5^6 .30

A = 30( 1 + 5 + ...+ 5^ 6) chia hết cho 30 

A chia hết cho 30

\(A=5+5^2+5^3+5^4+5^5+5^6+5^7+5^8\)

\(=\left(5+5^2\right)+\left(5^3+5^4\right)+\left(5^5+5^6\right)+\left(5^7+5^8\right)\)

\(=\left(5+5^2\right)+5^2\left(5+5^2\right)+5^4\left(5+5^2\right)+5^6\left(5+5^2\right)\)

\(=30+5^2.30+5^4.30+5^6.30\)

\(=30\left(1+5^2+5^4+5^6\right)⋮30\)

\(\Rightarrow A\) là bội của 30 (đpcm)

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

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

\(\Rightarrow A=\left(5+5^2\right)+5^2\left(5+5^2\right)+...+5^{28}\left(5+5^2\right)\)

\(\Rightarrow A=\left(5+5^2\right)\left(1++5^2+...+5^{28}\right)\)

\(\Rightarrow A=30\left(1++5^2+...+5^{28}\right)⋮30\)

dê mà, thôi mik giải cho k mik vs nha

A = 5 + 5^2 + 5^3 + .......... + 5^8

5A = 5^2 + 5^3 + 5^4 + .................. + 5^9 

5A - A =  5^2 + 5^3 + 5^4 + .................. + 5^9 - 5 - 5^2 - 5^3 -  .......... - 5^8

4A = 5^9 - 5

Suy ra A = ( 5^9 - 5 ) : 4 = 488280 chia hết cho 30

đừng quên k nha

dễ thế cũng hỏi,ko bít động não hả

kb voi tui di roi tui noi cach lam

ko biet lam

mk xin loi ket ban nha

Ta có: $A=\left(5+5^2\right)+\left(5^3+5^4\right)+\left(5^5+5^6\right)+\left(5^7+5^8\right)$

= $\left(5+5^2\right)+5^2\left(5+5^2\right)+5^4\left(5+5^2\right)+5^6\left(5+5^2\right)$

= $30+5^2.30+5^4.30+5^6.30$

= $30.(1+5^2+5^4+5^6)⋮30$

Vậy A là bội của 30 ok nha bn

\(\overline{ababab}=\overline{ab}.10101\)

Do  \(\overline{ab}⋮\overline{ab}\)

\(\Rightarrow\overline{ab}.10101⋮\overline{ab}\)

\(\Rightarrow\overline{ababab}⋮\overline{ab}\)