\(A=7+7^2+7^3+...+7^8\\=(7+7^2)+(7^3+7^4)+...+(7^7+7^8)\\=7\cdot(1+7)+7^3\cdot(1+7)+...+7^7\cdot(1+7)\\=7\cdot8+7^3\cdot8+...+7^7\cdot8\\=8\cdot(7+7^3+...+7^7)\)

Vì \(8\cdot(7+7^3+...+7^7)\vdots8\)

nên \(A\vdots8\)

\(A=7+7^2+7^3+...+7^8\)

\(A=\left(7+7^2\right)+\left(7^3+7^4\right)+...+\left(7^7+7^8\right)\)

\(A=56+7^2.\left(7+7^2\right)+...+7^6.\left(7+7^2\right)\)

\(A=56+7^2.56+...+7^6.56\)

\(A=56.\left(1+7^2+...+7^6\right)\)

Vì \(56⋮8\) nên \(56.\left(1+7^2+...+7^6\right)⋮8\)

Vậy \(A⋮8\)

\(#WendyDang\)

1.

\(A=7+7^2+7^3+...+7^{78}\)

\(=\left(7+7^2\right)+\left(7^3+7^4\right)+...+\left(7^{77}+7^{78}\right)\)

\(=7\left(1+7\right)+7^3\left(1+7\right)+...+7^{77}\left(1+7\right)\)

\(=7\cdot8+7^3\cdot8+...+7^{77}\cdot8\)

\(=\left(7+7^3+...+7^{77}\right)\cdot8\) chia hết cho 8

Vậy A chia hết cho 8 (đpcm)

\(A=3+3^2+3^3+...+3^{155}\)

\(=\left(3+3^2+3^3+3^4+3^5\right)+...+\left(3^{151}+3^{152}+3^{153}+3^{154}+3^{155}\right)\)

\(=3\left(1+3+3^2+3^3+3^4\right)+...+3^{151}\left(1+3+3^2+3^3+3^4\right)\)

\(=\left(3+...+3^{151}\right)\cdot121\) chia hết cho 121

Vậy A chia hết cho 121 (đpcm)

Ta có

a= 7(1+7)+7^3(1+7)+...+7^77(1+7) 

= 7.8 +7^3.8+...+7^77.8 

=8(7+7^3+...+7^77) chia hết cho 8 

=> a chia hết cho 8

a = 7 + 7^2 + 7^2 + 7^3 + 7^4 + ... + 7^78

a = ( 7 + 7^2 ) + ( 7^3 + 7^4 ) + ... + ( 7^77 + 7^78 )

a = 7( 1 + 7 ) + 7^3( 1 + 7 ) + ... + 7^77( 1 + 7 )

a = 7 . 8 + 7^3 . 8 + ... + 7^77 . 8

a = 8( 7 + 7^3 + ... + 7^77 )

=> a chia hết cho 8

A = 7+7² + 7³ +....+ 7¹⁰⁰

    = (7+7²) + (7³ + 7⁴)+.....+(7⁹⁹+7¹⁰⁰⁾

     = 7(1+7) + 7³(1+7)+.......+7⁹⁹(1+7)

    = 8(7+7²+7³+.....+ 7⁹⁹) chia hết cho 8  

A = 7 + 7² + 7³ + ... + 7⁹⁹ + 7¹⁰⁰

A = ( 7 + 7² ) + ( 7³ + 7⁴ ) + ... + ( 7⁹⁹ + 7¹⁰⁰ )

A = ( 1 + 7 ) . 7 + ( 1 + 7 ) . 7³ + ... + ( 1 + 7 ) . 7⁹⁹

A = 8 . 7 + 8 . 7³ + ... + 8 . 7⁹⁹

A = 8 . ( 7 + 7³ + ... + 7⁹⁹ )

=> A chia hết cho 8 (đpcm)

          giai

(7+7^2)+(7^3+7^4) + .................+(7^99+7^100)

7*(1+7) +7^3*(1+7)+.........+7^99*(1+7)

=8*(7+7^2+.........+7^99)

vi 8 nhan voi may cung chia het cho 8

=> 8*(7+7^2+............+7^99) chia het cho 8

phải là :

A= \(7+7^2+7^3+...+7^{99}+7^{100}\)

\(=\left(7+7^2\right)+\left(7^3+7^4\right)+...+\left(7^{99}+7^{100}\right)\)

\(=7.\left(1+7\right)+7^3.\left(1+7\right)+...+7^{99}.\left(1+7\right)\)

\(=7.8+7^3.8+...+7^{99}.8\\ =8.\left(7+7^3+7^{99}\right)\\ \Rightarrow A⋮8\)

Vậy \(A⋮8\)

Cho A=7^1+7^2+7^3+7^4+...+7^99+7^100. Chứng tỏ A chia hết cho 8.

\(A=1+4+4^2+...+4^{2012}=\left(1+4+4^2\right)+4^3\left(1+4+4^2\right)+...+4^{2010}\left(1+4+4^2\right)\)

\(=21+21.4^3+...+21.4^{2010}=21\left(1+4^3+...+4^{2010}\right)⋮21\)

\(B=1+7+7^2+...+7^{101}=\left(1+7\right)+7^2\left(1+7\right)+...+7^{100}\left(1+7\right)\)

\(=8+7^2.8+...+7^{100}.8=8\left(1+7^2+...+7^{100}\right)⋮8\)

Có \(A=7^1+7^2+7^3+...+7^{99}+7^{100}=\left(7^1+7^2\right)+\left(7^3+7^4\right)+...\left(7^{99}+7^{100}\right)\)

\(\Leftrightarrow A=7\left(1+7\right)+7^3\left(1+7\right)+...+7^{99}\left(1+7\right)=7.8+7^3.8+...+7^{99}.8=8\left(7+7^3+...+7^{99}\right)\)

Vì \(8\left(7+7^3+...+7^{99}\right)\)chia hết cho 8 nên \(A\)chia hết cho 8 (ĐPCM)

  __cho_mình_nha_chúc_bạn_học _giỏi__ 

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

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

\(A=7\cdot\left(7+7^2\right)+7^2\cdot\left(1+7^2\right)+7^5\cdot\left(1+7^2\right)+7^6\cdot\left(1+7^2\right)\)

\(A=7\cdot50+7^2\cdot50+7^5\cdot50+7^6\cdot50\)

\(A=50\cdot\left(7+7^2+7^5+7^6\right)\)

\(A=5\cdot10\cdot\left(7+7^2+7^5+7^6\right)\)

Ta có: 5 ⋮ 5

⇒ \(A=5\cdot10\cdot\left(7+7^2+7^5+7^6\right)\) ⋮ 5 (đpcm) 

A = 7 + 72 + 73 + 74 + 75 + 76 + 77 + 78

A =  (7 + 73) + (72+ 74) + (75 + 77) + (76 + 78)

A = 7.(1 + 72)  + 72.(1 + 72) + 75.(1 + 72) + 76.(1 + 72)

A = 7.( 1 + 49) + 72.( 1 + 49) + 75.(1 + 49) + 76. (1 + 49)

A = 7.50 + 72.50 + 75.40 + 76.50

A = 50.(7 + 72 + 75 + 76)

Vì 50 ⋮ 5 nên A = 50.(7 + 72 + 76) ⋮ 5 đpcm

A = 7 + 7² + 7³ + 7⁴ + 7⁵ + 7⁶ + 7⁷ + 7⁸

A =  (7 + 7³) + (7²+ 7⁴) + (7⁵ + 7⁷) + (7⁶ + 7⁸)

A = 7.(1 + 7²)  + 7².(1 + 7²) + 7⁵.(1 + 7²) + 7⁶.(1 + 7²)

A = 7.( 1 + 49) + 7².( 1 + 49) + 7⁵.(1 + 49) + 7⁶. (1 + 49)

A = 7.50 + 7².50 + 7⁵.50 + 7⁶.50

A = 50.(7 + 7² + 7⁵ + 7⁶)

Vì 50 ⋮ 5 nên A = 50.(7 + 7² + 7⁶) ⋮ 5 đpcm