A = 2 + 2² + 2³ + ... + 2⁶⁰

= (2 + 2²) + (2³ + 2⁴) + ... + (2⁵⁹ + 2⁶⁰)

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

= 2.3 + 2³.3 + ... + 2⁵⁹.3

= 3.(2 + 2³ + ... + 2⁵⁹) ⋮ 3

Vậy A ⋮ 3

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A = 2 + 2² + 2³ + ... + 2⁶⁰

= (2 + 2² + 2³) + (2⁴ + 2⁵ + 2⁶) + ... + (2⁵⁸ + 2⁵⁹ + 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 A ⋮ 7

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A = 2 + 2² + 2³ + ... + 2⁶⁰

= (2 + 2² + 2³ + 2⁴) + (2⁵ + 2⁶ + 2⁷ + 2⁸) + ... + (2⁵⁷ + 2⁵⁸ + 2⁵⁹ + 2⁶⁰)

= 30 + 2⁴.(2 + 2² + 2³ + 2⁴) + ... + 2⁵⁶.(2 + 2² + 2³ + 2⁴)

= 30.(1 + 2⁴ + ... + 2⁵⁶)

= 5.6.(1 + 2⁴ + ... + 2⁵⁶) ⋮ 5

Vậy A ⋮ 5

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

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

\(A=6+2^2.\left(2+2^2\right)+...+2^{58}.\left(2+2^2\right)\)

\(A=6+2^2.6+...+2^{58}.6\)

\(A=6.\left(1+2^2+...+2^{58}\right)\) 

Vì \(6⋮3\) nên \(6.\left(1+2^2+...+2^{58}\right)⋮3\)

Vậy \(A⋮3\)

___________

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

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

\(A=14+...+2^{57}.\left(2+2^2+2^3\right)\)

\(A=14+...+2^{57}.14\)

\(A=14.\left(1+...+2^{57}\right)\)

Vì \(14⋮7\) nên \(14.\left(1+...2^{57}\right)⋮7\)

Vậy \(A⋮7\)

____________

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

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

\(A=30+...+2^{56}.\left(2+2^2+2^3+2^4\right)\)

\(A=30+...+2^{56}.30\)

\(A=30.\left(1+...+2^{56}\right)\)

Vì \(30⋮5\) nên \(30.\left(1+...+2^{56}\right)⋮5\)

Vậy \(A⋮7\)

\(#WendyDang\)

A=2(1+2)+2^3(1+2)+...+2^59(1+2)

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

A=3(2+2^3+...+2^59) chia hết cho 3

Vậy a chia hết cho 3

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

A=2.7+...+2^58.7

A=7.(2+..+2^58) chia hết cho7

Vậy A chia hết cho 7

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

A=2.15+...+2^57.15

A=15.(2+...+2^57) chia hết cho 15 

Vậy A chia hết cho 15 

Vậy A chia hết cho 3,7,15

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

\(\Leftrightarrow A=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+.........+\left(2^{57}+2^{58}+2^{59}+2^{60}\right)\)

\(\Leftrightarrow A=2\left(1+2+2^2+2^3\right)+2^5\left(1+2+2^2+2^3\right)+..........+2^{57}\left(1+2+2^2+2^3\right)\)

\(\Leftrightarrow A=2.15+2^5.15+........+2^{57}.15\)

\(\Leftrightarrow A=15\left(2+2^5+......+2^{57}\right)⋮15\)

\(\Leftrightarrow A⋮15\left(đpcm\right)\)

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

\(\Rightarrow A=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+...+\left(2^{57}+2^{58}+2^{59}+2^{60}\right)\)

\(\Rightarrow A=2.\left(1+2+2^2+2^3\right)+2^5.\left(1+2+2^2+2^3\right)+...+2^{57}.\left(1+2+2^2+2^3\right)\)

\(\Rightarrow A=2.15+2^5.15+...+2^{57}.15\)

\(\Rightarrow A=15.\left(2+2^5+...+2^{57}\right)\)

\(\Rightarrow A⋮15\)

A=2+2²+...+2⁵⁹+2⁶⁰

=> A=2(1+2)+...+2⁵⁹(1+2)

        =2.3+...+2⁵⁹.3 chia het cho 3

2 cai kia ban lam tuong tu

⇔ 

⇔ 

⇔ 

⇔ 

⇒ A⋮ 3

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⇔ 

⇔ 

⇔ 

⇒ A⋮ 7

Hiện tại mình chưa tìm ra sao chia hết cho 5 nên bạn tự làm nhé cảm ơn bạn

a/ \(A=2+2^2+2^3+.....+2^{60}\)

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

\(=2\left(1+2\right)+2^3\left(1+2\right)+....+2^{59}\left(1+2\right)\)

\(=2.3+2^3.3+......+2^{59}.3\)

\(=3\left(2+2^3+....+2^{59}\right)⋮3\left(đpcm\right)\)

b/Ta có :

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

\(=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{58}+2^{59}+2^{60}\right)\)

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

\(=2.7+2^3.7+......+2^{58}.7\)

\(=7\left(2+2^3+.....+2^{58}\right)⋮7\left(đpcm\right)\)

c/ \(A=2+2^2+2^3+....+2^{60}\)

\(=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+....+\left(2^{57}+2^{58}+2^{59}+2^{60}\right)\)

\(=2\left(1+2+2^2+2^3\right)+2^5\left(1+2+2^2+2^3\right)+....+2^{57}\left(1+2+2^2+2^3\right)\)

\(=2.15+2^5.15+......+2^{57}.15\)

\(=15\left(2+2^5+......+2^{57}\right)⋮15\left(đpcm\right)\)

\(A=2+2^2+...+2^{59}+2^{60}\)\(=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{58}\left(1+2+2^2\right)\)

   \(=2.7+2^4.7+...+2^{58}.7\) \(=7\left(2+2^4+...+2^{58}\right)\) 

Dễ thấy A chia hết cho 7 (vì 7 chia hết cho 7)

Bài nãy đễ quá đi à

Ta có: \(A=2+2^2+2^3+...+2^{60}\)

\(\Rightarrow2A=2^2+2^3+2^4+...+2^{61}\)

\(A=2A-A=\left(2^2+...+2^{61}\right)-\left(2+...+2^{60}\right)\)

\(\Rightarrow A=2^{61}-2\)

b) \(2^4=...6\)

\(\Rightarrow2^{60}=...6\Rightarrow2^{61}-2=2^{60}.2-2=...0\)