Ta có: A = (2 + 2² + 2³) + (2⁴ + 2⁵ + 2⁶) + ..........+ (2⁵⁸ + 2⁵⁹ + 2⁶⁰)

             = 2 . (1 + 2 + 4 ) + 2⁴.(1+2+4) + ....... + 2⁵⁸.(1+2+4)

             = 2.7 + 2⁴.7 + .........+2⁵⁸.7

             = 7.(2+2⁴+.....+2⁵⁸)

Giải:

\(A=\text{( }2^1+2^2+2^3\text{)}+\left(2^4+2^5+2^6\right)+...+\left(2^{58}+2^{59}+2^{60}\right)\)

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

\(A=2.7+2^4.7+...+2^{58}.7\)

\(A=7.\left(2+2^4+2^{58}\right)⋮7\)

\(\Rightarrow A=2^1+2^2+2^3+2^4+....+2^{59}+2^{60}\) chia hết cho \(7\)

\(\Rightarrow A=\left(2^1+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+....+\left(2^{58}+2^{59}+2^{60}\right)\)

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

\(\Rightarrow A=2^1.7+2^4.7+...+2^{58}.7\)

\(\Rightarrow A=7\left(2^1+2^4+...+2^{58}\right)\)

\(\Rightarrow\)A chia hết cho 7 vì tích có chứ thừa số 7

Vậy A chia hết cho 7

\(A=2^1+2^2+2^3+.....+2^{59}+2^{60}\\ =\left(2+2^2+2^3\right)+...+\left(2^{58}+2^{59}+2^{60}\right)\\ =2\left(1+2+2^2\right)+....+2^{58}\left(1+2+2^2\right)\\ =7\left(2+....+2^{58}\right)⋮7\)

A=2+2^2+2^3+...+2^59+2^60(có 60 số hạng)

A=(2+2^2+2^3)+(2^4+2^5+2^6)+...+(2^58+2^59+2^60)[có 20 nhóm]

A=14*1+2^3*(2+2^2+2^3)+...+2^57*(2+2^2+2^3)

A=14*1+2^3*14+...+2^57*14

A=14*(1+2^3+...+2^57)

A=7*2*(1+2^3+...+2^57) chia hết cho 7(tick nha)

THANK NHÌU NHA

NHƯNG TỚ CHƯA HIÊU DẤU *

GIẢI THICK VỚI

 A= (2¹+2²+2³)+(2⁴+2⁵+2⁶)+...+(2⁵⁸+2⁵⁹+2⁶⁰)

   =2⁰(2¹+2²+2³)+2³(2¹+2²+2³)+...+2⁵⁷(2¹+2²+2³)

   =(2¹+2²+2³)(2⁰+2³+...+2⁵⁷⁾

   =     14(2⁰+2³+...+2⁵⁷) chia hết cho 7

Vậy A chia hết cho 7     

gọi 1/41+1/42+1/43+...+1/80=S

ta có :

S>1/60+1/60+1/60+...+1/60

S>1/60 x 40

S>8/12>7/12

Vậy S>7/12

cho mình hỏi nhờ cũng cái đề bài này nhưng chia hết cho 37 làm thế nào

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.

2+2^2+...+2^60

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

=6.(1+2^2+...+2^58)

=3.2(1+2^2+...+2^58)chia hết cho 3

\(A=2^1+2^2+2^3+...+2^{60}\)

\(=\left(2^1+2^2+2^3\right)+...+\left(2^{58}+2^{59}+2^{60}\right)\)

\(=\left(2.1+2.2+2.2^2\right)+...+\left(2^{58}.1+2^{58}.2+2^{58}.2^2\right)\)

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

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

\(=\left(2+2^{58}\right).7⋮7\)hay \(A⋮7\)

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

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

A=2.7+...+2^58.7

A=7(2+2^4+....+2^58) chia hết cho 7

vậy...

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) chia hết cho 3

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) chia hết cho 7

A = 2 + 2² + 2³ + ... + 2⁶⁰

= (2 + 2² + 2³ + 2⁴) + (2⁵ + 2⁶ + 2⁷ + 2⁸) + ... + (2⁵⁷ + 2⁵⁸ + 2⁵⁹ + 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) chia hết cho 15

bạn có nhầm đề ko ? Xem lại 2⁵⁰ +2⁶⁰ 

a)A= 2+2^2+2^3+...+2^59+2^60

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

A=2.(1+2)+2^3.(1+2)+2^5.(1+2)+...+2^59.(1+2)

A=2.3+2^3.3+2^5.3+...+2^59.3

A=3.(2+2^3+2^5+...+2^59)

=>A chia hết cho 3

Vậy A chia hết cho 3

b)A= 2+2^2+2^3+..........................+2^59+2^60

A=(2+2^2+2^3)+(2^4+2^5+2^6)+(2^7+2^8+2^9)+...+(2^58+2^59+2^60)

A=2.(1+2+2^2)+2^4.(1+2+2^2)+2^7.(1+2+2^2)+...+2^58.(1+2+2^2)

A=2.7+2^4.7+2^7.7+...+2^58.7

A=7.(2+2^4+2^7+...+2^58)

=>A chia hết cho 7

Vậy A chia hết cho 7

c)A= 2+2^2+2^3+..........................+2^59+2^60

A=(2+2^2+2^3+2^4)+(2^5+2^6+2^7+2^8)+(2^9+2^10+2^11+2^12)+...+(2^57+2^58+2^59+2^60)

A=2.(1+2+2^2+2^3)+2^5.(1+2+2^2+2^3)+2^9.(1+2+2^2+2^3)+...+2^57.(1+2+2^2+2^3)

A=2.15+2^5.15+2^9.15+...+2657.15

A=15.(2+2^5+2^9+...+2^57)

=>A chia hết cho 15

Vậy A chia hết cho15

A=2^1+2^2+...+2^60=(2^1+2^2+2^3)+...+(2^58+2^59+2^60)=2^1.(1+2+2^2)+2^4.(1+2+2^2)+...+2^58.(1+2+2^2)=2^1.7+2^4.7+...+2^58.7

  =7.(2^1+2^4+...+2^58) chia hết cho 7 (đpcm)

A = 2 ^ 1 + 2 ^ 2 + ... + 2 ^ 59 + 2 ^ 60

A = ( 2 ^ 1 + 2 ^ 2 + 2 ^ 3 ) + ... + ( 2 ^ 58 + 2 ^ 59 + 2 ^ 60 )

A = ( 2 ^ 1 + 2 ^ 2 + 2 ^ 3 ) + ... + ( 2 ^ 1 + 2 ^ 2 + 2 ^ 3 ) . 2 ⁵⁷

A = 14 + ... + 14 . 2 ⁵⁷

A = 14 ( 1 + ... + 2 ⁵⁷ )

Vì 14 chi hết cho 7

=> A chia hết cho  7

A=(2¹+2²+2³)+(2⁴+2⁵+2⁶)+...+(2⁵⁸+2⁵⁹+2⁶⁰)

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

=>A=2.7+2⁴.7+...+2⁵⁸.7

=>A=7.(2+2⁴+...+2⁵⁸) chia hết  cho 7

=>A chia hết cho 7 (đpcm)