Sai đề ruj A=137256 ko thể chia hết cho 50

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ta có: 7+7^2+7^3+... + 7^8

=( 7+7^2) +( 7^3 +7^4)+...+(7^7 +7^8)

= 50 + 7^2(7+7^2)+...+ 7^6(7+ 7^2)

= 50 + 7^2 . 50+...+ 7^6 . 50

= 50.( 1+7^2 + ... + 7^6) chia hết cho 50

Vậy 7 + 7^2 + 7^3 + 7^4 + 7^5 +7^6 +7^7 +7^8 chia hết cho 50

k cho mk nha

\(7+7^2+7^3+7^4+7^5+7^6\)

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

\(=7\left(1+7\right)+7^3\left(1+7\right)+7^5\left(1+7\right)\)

\(=8\left(7+7^3+7^5\right)\)\(⋮8\)(điều phải chứng minh)

cảm ơn Khải nhé

a, 11 + 11² + 11³ + ... + 11⁷ + 11⁸

= (11 + 11²) + (11³ + 11⁴) + ... + (11⁷ + 11⁸)

= 11(1 + 11) + 11³(1 + 11) + ... + 11⁷(1 + 11)

= 11.12 + 11³.12 + .... + 11⁷.12

= 12(11 + 11³ + ... + 11⁷) chia hết cho 12

b, 7 + 7² + 7³ + 7⁴

= (7 + 7³) + (7² + 7⁴)

= 7(1 + 7²) + 7²(1 + 7²)

= 7.50 + 7².50

= 50(7  + 7²) chia hết cho 50

c, 3 + 3² + 3³ + 3⁴ + 3⁵ + 3⁶

= (3 + 3² + 3³) + (3⁴ + 3⁵ + 3⁶)

= 3(1 + 3 + 3²) + 3⁴(1 + 3 + 3²)

= 3.13 + 3⁴.13

= 13(3 + 3⁴) chia hết cho 13

 A= 7+7²+7³+....+7⁵⁰ 

    = (7 + 7³ ) + (7² + 7⁴) + ..... + (7⁴⁷ + 7⁴⁹) + (7⁴⁸ + 7⁵⁰)

    = 7.(1 + 49) + 7².(1 + 49) + ...... + 7⁴⁷.(1 + 49) + 7⁴⁸.(1 + 49) 

   = 7. 50 + 7².50 + ...... +  7⁴⁷.50 + 7⁴⁸.50

  = 50.( 7 + 7² + ..... +7⁴⁷ + 7⁴⁸) chai hết 50 ( đpcm)

\(7^6+7^5-7^4=7^4\left(7^2+7^1-1\right)=7^4\left(49+7-1\right)=55⋮11\)

\(\Rightarrow7^6+7^5-7^4⋮11\)

So sánh hay chứng minh vậy bạn

Ta có: \(A=7^3+7^4+7^5+7^6+...+7^{98}\)

\(\Rightarrow A=\left(7^3+7^4\right)+\left(7^5+7^6\right)+...+\left(7^{97}+7^{98}\right)\)

\(\Rightarrow A=7^3\left(1+7\right)+7^5\left(1+7\right)+...+7^{97}\left(1+7\right)\)

\(\Rightarrow A=7^3.8+7^5.8+...+7^{97}.8\)

\(\Rightarrow A=\left(7^3+7^5+...+7^{97}\right).8⋮8\)

\(\Rightarrow A⋮8\)

Vậy \(A⋮8\)

A = 7³ + 7⁴ + 7⁵ + 7⁶ + ... + 7⁹⁷ + 7⁹⁸

A = (7³ + 7⁴) + (7⁵ + 7⁶) + ... + (7⁹⁷ + 7⁹⁸)

A = 7³.(1 + 7) + 7⁵.(1 + 7) + ... + 7⁹⁷.(1 + 7)

A = 7³.8 + 7⁵.8 + ... + 7⁹⁷.8

A = 8.(7³ + 7⁵ + ... + 7⁹⁷)

=> A ⋮ 8.