Úi gời cơi cộng chấm chấm chấm :)))

+ Ta có: \(A=2+2^2+2^3+2^4+...+2^{2010}\)

\(A=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{2009}\left(1+2\right)\)

\(A=2.3+2^3.3+...+2^{2009}.3\)

\(A=3\left(2+2^3+...+2^{2010}\right)⋮3\)

-> Đpcm

+ Ta có: \(A=2+2^2+2^3+2^4+...+2^{2010}\)

\(A=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+....+2^{2008}\left(1+2+2^2\right)\)

\(A=2.7+2^4.7+...+2^{2008}.7\)

\(A=7\left(2+2^4+...+2^{2008}\right)⋮7\)

-> Đpcm

\(A=2^1+2^2+...+2^{2010}\)

\(=\left(2^1+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{2009}+2^{2010}\right)\)

\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{2009}\left(1+2\right)\)

\(=3\left(2+2^3+...+2^{2009}\right)⋮3\)

\(A=2+2^2+2^3+...+2^{2010}\)

\(=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{2008}+2^{2009}+2^{2010}\right)\)

\(=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{2008}\left(1+2+2^2\right)\)

\(=7\left(2+2^4+...+2^{2008}\right)⋮7\)

 + Chứng minh chia hết cho 3  
 \(A=2^1+2^2+2^3+2^4+...+2^{2010}\)
\(=\left(2^1+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{2009}+2^{2010}\right)\)
\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{2009}\left(1+2\right)\)
\(=\left(1+2\right)\left(2+2^3+...+2^{2009}\right)\)
\(=3\left(2+2^3+...+2^{2009}\right)\)
Vì \(3\) ⋮ \(3\)
⇒ \(A\) ⋮ \(3\)



+ Chứng minh chia hết cho 7
\(A=2^1+2^2+2^3+2^4+...+2^{2010}\)
\(=\left(2^1+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{2008}+2^{2009}+2^{2010}\right)\)
\(=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{2008}\left(1+2+2^2\right)\)
\(=\left(1+2+2^2\right)\left(2+2^4+...+2^{2008}\right)\)
\(=7\left(2+2^4+...+2^{2008}\right)\)
Vì \(7\) ⋮ \(7\)
⇒ \(A\) ⋮ \(7\)

A=2\(^1\)+2\(^2\)+...+2\(^{2010}\)

=(2\(^1\)+2\(^2\))+(2\(^3\)+2\(^4\))+...+(2\(^{2009}\)+2\(^{2010}\))

=2(1+2)+2\(^3\)(1+2)+...+2\(^{2009}\)(1+2)

=3(2+2\(^3\)+...+2\(^{2009}\))⋮3

... tìm số dư khi chia hết???

nếu nó chia hết thì số dư bằng 0 rồi

bạn nếu cách làm đi

Trời trời, mình làm cho bạn câu khi nãy bạn phải biết vận dụng cho mấy bài sau chứ, câu này giống i lột câu khi nãy luôn ấy, nhưng thôi, khá rảnh nên:vv

+Ta có: \(B=3+3^2+3^3+3^4+...+3^{2010}\)

-> \(B=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{2009}\left(1+3\right)\)

-> \(B=3.4+3^3.4+...+3^{2009}.4\)

-> \(B=4\left(3+3^3+...+3^{2009}\right)⋮4\)

-> Đpcm 

+ Ta có: \(B=3+3^2+3^3+3^4+....+3^{2010}\)

-> \(B=3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+...+3^{2008}\left(1+3+3^2\right)\)

-> \(B=3.13+3^4.13+...+.3^{2008}.13\)

-> \(B=13\left(3+3^4+...+3^{2008}\right)⋮13\)

-> Đpcm

Ta có: \(B=3^1+3^2+3^3+3^4+...+3^{2010}\)

\(=3^1\cdot\left(1+3\right)+3^3\cdot\left(1+3\right)+...+3^{2009}\cdot\left(1+3\right)\)

\(=\left(1+3\right)\cdot\left(3^1+3^3+...+3^{2009}\right)\)

\(=4\cdot\left(3+3^3+...+3^{2009}\right)⋮4\)(đpcm)

Ta có: \(B=3^1+3^2+3^3+3^4+...+3^{2010}\)

\(=3\left(1+3+3^2\right)+3^4\cdot\left(1+3+3^2\right)+...+3^{2008}\cdot\left(1+3+3^2\right)\)

\(=\left(1+3+3^2\right)\cdot\left(3+3^4+...+3^{2008}\right)\)

\(=13\cdot\left(3+3^4+...+3^{2008}\right)⋮13\)(đpcm)

A = 3 + 3² + 3³ + ... + 3¹⁰⁰  

Số số hạng của A = ( 100 - 1 ) : 1 + 1 = 100 ssh . Ta chia A thanh 25 nhóm , mỗi nhóm cs 4 ssh . 

=> A = ( 3 + 3² + 3³ + 3⁴ ) + .... + ( 3⁹⁷ + 3⁹⁸ + 3⁹⁹ + 3¹⁰⁰ ) 

     A  =  3. ( 1 + 3 + 3² + 3³ ) + .... + 3⁹⁷.( 1 + 3 + 3² + 3³ ) 

     A  = 3. 40 + ... + 3⁹⁷ . 40 

    A  = 40. ( 3 + ... + 3⁹⁷ ) 

   =>  A \(⋮\) 40 ( đpcm ) 

A = 3 + 32 + 33 + ... + 3100  

Số số hạng của A = ( 100 - 1 ) : 1 + 1 = 100 ssh . Ta chia A thanh 25 nhóm , mỗi nhóm cs 4 ssh . 

=> A = ( 3 + 32 + 33 + 34 ) + .... + ( 397 + 398 + 399 + 3100 ) 

     A  =  3. ( 1 + 3 + 32 + 33 ) + .... + 397.( 1 + 3 + 32 + 33 ) 

     A  = 3. 40 + ... + 397 . 40 

    A  = 40. ( 3 + ... + 397 ) 

   =>  A $⋮$ 40 ( đpcm ) 

HT

A = (3+3^2+3^3+3^4)+(3^5+3^6+3^7+3^8)+.....+(3^97+3^98+3^99+3^100)

   = 120+3^4.(3+3^2+3^3+3^4)+.....+3^96.(3+3^2+3^3+3^4)

   = 120+3^4.110+....+3^96.120

   = 120.(1+3^4+.....+3^96) chia hết cho 120

=> ĐPCM

Tk mk nha

dư 0 nhé

Ta có: ( Sửa đề )

\(A=4+4^2+4^3+...+4^{2021}+4^{2022}\)

\(A=\left(4+4^2\right)+\left(4^3+4^4\right)+...+\left(4^{2021}+4^{2022}\right)\)

\(A=20+4^2.\left(4+4^2\right)+...+4^{2020}.\left(4+4^2\right)\)

\(A=20+4^2.20+...+4^{2020}.20\)

\(A=20.\left(1+4^2+...+4^{2020}\right)\)

Vì \(20⋮20\) nên \(20.\left(1+4^2+...+4^{2020}\right)\)

Vậy \(A⋮20\)

\(#WendyDang\)