mình ko biết

phải là chứng minh A chia hết cho 121

\(A=3+3^2+3^3+...+3^{2012}\\ A=\left(3+3^2+3^3+3^4\right)+...+\left(3^{2009}+3^{2010}+3^{2011}+3^{2012}\right)\\ A=120+...+3^{2008}.120\\ A=120.\left(1+...+3^{2008}\right)⋮120\)

\(=\left(3+3^2+3^3+3^4\right)+...+3^{2008}\left(3+3^2+3^3+3^4\right)\)

\(=120+...+3^{2008}.120=120\left(1+...+3^{2008}\right)⋮120\)

\(A=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{89}+3^{90}\right)\\ A=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{89}\left(1+3\right)\\ A=3\cdot4+3^3\cdot4+...+3^{89}\cdot4\\ A=4\left(3+3^3+...+3^{89}\right)⋮4\)

A = ( 3 + 3 2 ) + ( 3 3 + 3 4 ) + . . . + ( 3 89 + 3 90 )

A = 3 ( 1 + 3 ) + 3 3 ( 1 + 3 ) + . . . + 3 89 ( 1 + 3 )

A = 3 ⋅ 4 + 3 3 ⋅ 4 + . . . + 3 89 ⋅ 4

A = 4 ( 3 + 3 3 + . . . + 3 89 ) ⋮ 4

đề sai rồi bạn

`#3107.101107`

\(A=1+3+3^2+3^3+...+3^{101}\)

$A = (1 + 3 + 3^2) + (3^3 + 3^4 + 3^5) + ... + (3^{99} + 3^{100} + 3^{101}$

$A = (1 + 3 + 3^2) + 3^3 (1 + 3 + 3^2)  + ... + 3^{99}(1 + 3 + 3^2)$

$A = (1 + 3 + 3^2)(1 + 3^3 + ... + 3^{99})$

$A = 13(1 + 3^3 + ... + 3^{99})$

Vì `13(1 + 3^3 + ... + 3^{99}) \vdots 13`

`\Rightarrow A \vdots 13`

Vậy, `A \vdots 13.`

\(A=1+3+3^2+3^3+3^4+3^5+...+3^{101}\\=(1+3+3^2)+(3^3+3^4+3^5)+(3^6+3^7+3^8)+...+(3^{99}+3^{100}+3^{101})\\=13+3^3\cdot(1+3+3^2)+3^6\cdot(1+3+3^2)+...+3^{99}\cdot(1+3+3^2)\\=13+3^3\cdot13+3^6\cdot13+...+3^{99}\cdot13\\=13\cdot(1+3^3+3^6+...+3^{99})\)

Vì \(13\cdot(1+3^3+3^6...+3^{99}\vdots13\)

nên \(A\vdots13\)

\(\text{#}Toru\)