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

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

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

a) \(A=2+2^2+2^3+\dots+2^{60}\)

\(2A=2^2+2^3+2^4+\dots+2^{61}\)

\(2A-A=\left(2^2+2^3+2^4+\dots+2^{61}\right)-\left(2+2^2+2^3+\dots+2^{60}\right)\)

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

Vậy: \(A=2^{61}-2\).

b)

+) \(A=2+2^2+2^3+\dots+2^{60}\)

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

\(=2\cdot\left(1+2\right)+2^3\cdot\left(1+2\right)+2^5\cdot\left(1+2\right)+\dots+2^{59}\cdot\left(1+2\right)\)

\(=2\cdot3+2^3\cdot3+2^5\cdot3+\dots+2^{59}\cdot3\)

\(=3\cdot\left(2+2^3+2^5+\dots+2^{59}\right)\)

Vì \(3\cdot\left(2+2^3+2^5+\dots+2^{59}\right)⋮3\) nên \(A⋮3\)

+) \(A=2+2^2+2^3+\dots+2^{60}\)

\(=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+\left(2^9+2^{10}+2^{11}+2^{12}\right)+\dots+\left(2^{57}+2^{58}+2^{59}+2^{60}\right)\)

\(=2\cdot\left(1+2+2^2+2^3\right)+2^5\cdot\left(1+2+2^2+2^3\right)+2^9\cdot\left(1+2+2^2+2^3\right)+\dots+2^{57}\cdot\left(1+2+2^2+2^3\right)\)

\(=2\cdot15+2^5\cdot15+2^9\cdot15+\dots+2^{57}\cdot15\)

\(=15\cdot\left(2+2^5+2^9+\dots+2^{57}\right)\)

Vì \(15⋮5\) nên \(15\cdot\left(2+2^5+2^9+\dots+2^{57}\right)⋮5\)

hay \(A\vdots5\)

+) \(A=2+2^2+2^3+\dots+2^{60}\)

\(=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+\left(2^7+2^8+2^9\right)+\dots+\left(2^{58}+2^{59}+2^{60}\right)\)

\(=2\cdot\left(1+2+2^2\right)+2^4\cdot\left(1+2+2^2\right)+2^7\cdot\left(1+2+2^2\right)+\dots+2^{58}\cdot\left(1+2+2^2\right)\)

\(=2\cdot7+2^4\cdot7+2^7\cdot7+\dots+2^{58}\cdot7\)

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

Vì \(7\cdot\left(2+2^4+2^7+\dots+2^{58}\right)⋮7\) nên \(A⋮7\)

$Toru$

a) �=2+22+23+⋯+260A=2+22+23+⋯+260

2�=22+23+24+⋯+2612A=22+23+24+⋯+261

2�−�=(22+23+24+⋯+261)−(2+22+23+⋯+260)2A−A=(22+23+24+⋯+261)−(2+22+23+⋯+260)

�=261−2A=261−2

Vậy: �=261−2A=261−2.

b)

+) �=2+22+23+⋯+260A=2+22+23+⋯+260

=(2+22)+(23+24)+(25+26)+⋯+(259+260)=(2+22)+(23+24)+(25+26)+⋯+(259+260)

=2⋅(1+2)+23⋅(1+2)+25⋅(1+2)+⋯+259⋅(1+2)=2⋅(1+2)+23⋅(1+2)+25⋅(1+2)+⋯+259⋅(1+2)

=2⋅3+23⋅3+25⋅3+⋯+259⋅3=2⋅3+23⋅3+25⋅3+⋯+259⋅3

=3⋅(2+23+25+⋯+259)=3⋅(2+23+25+⋯+259)

Vì 3⋅(2+23+25+⋯+259)⋮33⋅(2+23+25+⋯+259)⋮3 nên $A⋮3$

+) �=2+22+23+⋯+260A=2+22+23+⋯+260

=(2+22+23+24)+(25+26+27+28)+(29+210+211+212)+⋯+(257+258+259+260)=(2+22+23+24)+(25+26+27+28)+(29+210+211+212)+⋯+(257+258+259+260)

=2⋅(1+2+22+23)+25⋅(1+2+22+23)+29⋅(1+2+22+23)+⋯+257⋅(1+2+22+23)=2⋅(1+2+22+23)+25⋅(1+2+22+23)+29⋅(1+2+22+23)+⋯+257⋅(1+2+22+23)

=2⋅15+25⋅15+29⋅15+⋯+257⋅15=2⋅15+25⋅15+29⋅15+⋯+257⋅15

=15⋅(2+25+29+⋯+257)=15⋅(2+25+29+⋯+257)

Vì $15⋮5$ nên 15⋅(2+25+29+⋯+257)⋮515⋅(2+25+29+⋯+257)⋮5

hay $A⋮5$

+) �=2+22+23+⋯+260A=2+22+23+⋯+260

=(2+22+23)+(24+25+26)+(27+28+29)+⋯+(258+259+260)=(2+22+23)+(24+25+26)+(27+28+29)+⋯+(258+259+260)

=2⋅(1+2+22)+24⋅(1+2+22)+27⋅(1+2+22)+⋯+258⋅(1+2+22)=2⋅(1+2+22)+24⋅(1+2+22)+27⋅(1+2+22)+⋯+258⋅(1+2+22)

=2⋅7+24⋅7+27⋅7+⋯+258⋅7=2⋅7+24⋅7+27⋅7+⋯+258⋅7

=7⋅(2+24+27+⋯+258)=7⋅(2+24+27+⋯+258)

Vì 7⋅(2+24+27+⋯+258)⋮77⋅(2+24+27+⋯+258)⋮7 nên $A⋮7$

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

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

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

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

TA có:VÌ 2= 2^1 

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

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

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

A= \(3.\left(2+2^3+...+2^{60}\right)\)chia hết cho 3

=) A chia hết cho3( đpcm)

Ta lại có:

A= \(2^1+2^2+2^3+...+2^{60}\)

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

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

A= \(7.\left(2+...+2^{58}\right)\)chia hết cho 7

=) A chia hết cho 7( đpcm)

ahihi

 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

Bài 3:

\(A=5+5^2+..+5^{12}\)

\(5A=5\cdot\left(5+5^2+..5^{12}\right)\)

\(5A=5^2+5^3+...+5^{13}\)

\(5A-A=\left(5^2+5^3+...+5^{13}\right)-\left(5+5^2+...+5^{12}\right)\)

\(4A=5^2+5^3+...+5^{13}-5-5^2-...-5^{12}\)

\(4A=5^{13}-5\)

\(A=\dfrac{5^{13}-5}{4}\)

A=2+2²+2³+....+2⁶⁰

A=(2+2²)+(2³+2⁴)+...+(2⁵⁹+2⁶⁰)

A=2.3+2³.3+....+2⁵⁹.3 chia hết cho 3 

2) A=2+2²+2³+...+2⁶⁰

A=(2+2²+2³)+.... +(2⁵⁸+2⁵⁹+2⁶⁰)

A=2.7+....+2⁵⁸.7 chia hết cho 7 

3) A=2+2²+2³+....+2⁶⁰

A=(2+2²+2³+2⁴)+....+(2⁵⁷+2⁵⁸+2⁵⁹+2⁶⁰)

A=2.15+....+2⁵⁷.15 chia hết cho 15

A= (2+2²)+(2³+2⁴)+...+(2⁵⁹+2⁶⁰)

A=2.(1+2)+2³.(1+2)+...+2⁵⁹.(1+2)

A=2.3+2³.3+...+2⁵⁹.3

A=3.(2+2³+...+2⁵⁹)

Vì 3 chia hết cho 3 => 3.(2+2³+...+2⁵⁹)  chia hết cho 3

=>A  chia hết cho 3

A= (2+2²+2³)+...+(2⁵⁸+2⁵⁹+2⁶⁰)

A=2.(1+2+2²)+...+2⁵⁸.(1+2+2²)

A=2.7+...+2⁵⁸.7

A=7.(2+...+2⁵⁸)

Vì 7  chia hết cho 7 =>7.(2+...+2⁵⁸)  chia hết cho 7

=>A  chia hết cho 7

A= (2+2²+2³+2⁴)+...+(2⁵⁷+2⁵⁸+2⁵⁹+2⁶⁰)

A=2.(1+2+2²+2³)+...+2⁵⁷.(1+2+2²+2³)

A=2.15 +...+2⁵⁷.15

A=15.(2+...+2⁵⁷)

vì 15 chia hết cho15=>15.(2+...+2⁵) chia hết cho 15

=>A chia hết cho 15