Question

Câu 3: Cho A = 7 + 7² + 7³ + ... + 7¹¹⁹ + 7¹²⁰. Chứng minh rằng A chia hết cho 57.

Answer 1

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

Answer 2

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.