7²⁰ + 49¹¹ + 343⁷ = ( 7¹ )²⁰ + ( 7² )¹¹ + ( 7³ )⁷ =7²⁰ + 7²¹ + 7²² = 7²⁰ ( 1 + 7 + 7² ) = 7²⁰.57

Vì 57 chia hết cho 57 nên 7²⁰ .57 chia hết cho 57 

=> 7²⁰ + 49¹¹ + 343⁷ chia hết cho 57 ( đpcm )

Ta có: 7^20 + 49^11 + 343^7 = 7^20 + (7^2)^11 + (7^3)^7 = 7^20 + 7^22 + 7^21 = 7^20(1 + 7 + 7^2) = 7^20.57 chia hết cho 57

=>ĐPCM

\(7^{20}+49^{11}+343^7\)

\(=7^{20}+\left(7^2\right)^{11}+\left(7^3\right)^7\)

\(=7^{20}+7^{22}+7^{21}\)

\(=7^{20}\left(1+7^2+7\right)\)

\(=7^{20}.57⋮57\)

\(\Leftrightarrowđpcm\)

Ta có: 7²⁰+49¹¹+343⁷= 7²⁰+(7²)¹¹+ (7³)⁷

= 7²⁰+7²²+7²¹=7²⁰.(1+7+7²)=7²⁰.57 chia hết cho 57

\(\Rightarrow\)7²⁰+49¹¹+343⁷ chia hết cho 57

7²⁰ + 49¹¹ + 343⁷

= 7²⁰ + (7²)¹¹ + (7³)⁷

= 7²⁰ + 7²² + 7²¹

= 7²⁰ + 7^{20 + 2} + 7^{20 + 1}

= 7²⁰ + 7²⁰.7² + 7²⁰.7

= 7²⁰(1 + 7² + 7)

= 7²⁰.57

Vì 57 ⋮ 57 nên 7²⁰.57 ⋮ 57

Hay 7²⁰ + 49¹¹ + 343⁷ ⋮ 57

$ 7^{20} + 49^{11} + 343^{7} \\ = 7^{20} + (7^{2})^{11} + (7^{3})^{7} \\ = 7^{20} + 7^{22} + 7^{21} \\ = 7^{20}(1 + 49 + 7) \\ = 7^{20} . 57 \vdots 57 $

\(A=7+7^2+7^3+...+7^{119}+7^{120}\)

\(\Rightarrow7A=7^2+7^3+7^4+...+7^{120}+7^{121}\)

\(\Rightarrow7A-A=\left(7^2+7^3+...+7^{120}+7^{121}\right)-\left(7+7^2+...+7^{119}+7^{120}\right)\)

\(\Rightarrow6A=7^2+7^3+...+7^{120}+7^{121}-7-7^2-...-7^{119}-7^{120}\)

\(\Rightarrow6A=7^{121}-7\)

\(\Rightarrow A=\dfrac{7^{121}-7}{6}\)

A=7+7²+7³+...+7²⁰¹⁶

=(7+7²)+(7³+7⁴)+...+(7²⁰¹⁵+7²⁰¹⁶)

=7.(1+7)+7³.(1+8)+...+7²⁰¹⁵.(1+7)

=7.8+7³.8+...+7²⁰¹⁵.8

=8.(7+7³+...+7²⁰¹⁵) chia hết cho 8 (đpcm)

A=7+7²+7³+...+7²⁰¹⁶

=(7+7²+7³)+...+(7²⁰¹⁴+7²⁰¹⁵+7²⁰¹⁶)

=7.(1+7+7²)+...+7²⁰¹⁴.(1+7+7²)

=7.57+...+7²⁰¹⁴.57

=57.(7+...+7²⁰¹⁴) chia hết cho 57 (đpcm)