A = 7 + 7² + 7³ + 7⁴ + 7⁵ + 7⁶ + ... + 7²¹

= (7 + 7² + 7³) + (7⁴ + 7⁵ + 7⁶) + ... + (7¹⁹ + 7²⁰ + 7²¹)

= 7.(1 + 7 + 7²) + 7⁴.(1 + 7 + 7²) + ... + 7¹⁹.(1 + 7 + 7²)

= 7.57 + 7⁴.57 + ... + 7¹⁹.57

= 57.(7 + 7⁴ + ... + 7¹⁹) ⋮ 57

Vậy A ⋮ 57

A = 7 + 7² + 7³ + 7⁴ + 7⁵ + 7⁶ + ... + 7²¹
A=(7 + 7² + 7³) + (7⁴ + 7⁵ + 7⁶) + ... + (7¹⁹ + 7²⁰ + 7²¹) 

A= 7.(1 + 7 + 7²) + 7⁴.(1 + 7 + 7²) + ... + 7¹⁹.(1 + 7 + 7²)

A= 7.57 + 7⁴.57 + ... + 7¹⁹.57

A= 57.(7 + 7⁴ + ... + 7¹⁹) ⋮ 57

  Do 57 ⋮ 57
=> Vậy A ⋮ 57

\(A=7\left(1+7+7^2\right)+7^4\left(1+7+7^2\right)+...+7^{118}\left(1+7+7^2\right)\)

\(=57\left(7+7^4+...+7^{118}\right)⋮57\)

\(A=7\left(1+7+7^2\right)+...+7^{118}\left(1+7+7^2\right)\)

\(=57\left(7+...+7^{118}\right)⋮57\)

A =7(1+7+7₂)+7⁴(1+7+7₂)+...+7¹¹⁸(1+7+7²)A=7(1+7+7²)+74(1+7+7²)+...+7¹¹⁸(1+7+7²)

=57 (7+7⁴+...+71¹⁸)⋮57

\(A=7\left(1+7+7^2\right)+7^4\left(1+7+7^2\right)+...+7^{118}\left(1+7+7^2\right)=7.57+7^4.57+...+7^{118}.57=57\left(7+7^4+...+7^{118}\right)⋮57\)

Lời giải:
$A=(7+7^2+7^3)+(7^4+7^5+7^6)+....+(7^{118}+7^{119}+7^{120})$
$=7(1+7+7^2)+7^4(1+7+7^2)+...+7^{118}(1+7+7^2)$

$=7.57+7^4.57+...+7^{118}.57$

$=57(7+7^4+...+7^{118})\vdots 57$ 

Ta có đpcm.

ý bạn là sao ? mình muốn cách giải cơ 

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

\(A=\left(7+7^2+7^3\right)+...+\left(7^{118}+7^{119}+7^{120}\right)\)

\(A=7\left(1+7+7^2\right)+...+7^{118}\left(1+7+7^2\right)\)

\(A=7.57+7^4.57+...+7^{118}.57\)

\(A=57\left(7+7^4+...+7^{118}\right)\)

\(\Rightarrow A⋮57\)

Sợ quá!

106 - 57 = (2.5)6 - 56.5 = 26.56 - 56.5=56.(26 - 5)=56.59⋮ 59

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)