C

A

B

A = 8⁸ + 2²⁰

= (2³)⁸ + 2²⁰

= 2²⁴ + 2²⁰

= 2²⁰.(2⁴ + 1)

= 2²⁰.17 ⋮ 17

Vậy A ⋮ 17

`#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 \(A\vdots13\)

\(\text{#}Toru\)

A=1+3+3^2+3^3+...+3^98+3^99+3^100

A=(1+3+ 3^2)+(3^3+3^4+3^5)+...+(3^98+3^99+3^100)

A=(1+3+3^2)+3^3x(1+3+3^2)+...+3^98x(1+3+3^2)

A=13x3^3x13+...+3^98x13

=> 13x(1+3+3^3+...+3^98)chia hết cho 13

Vậy A chia hết cho 13

câu b đâu bạn ?

cho A = 1 + 3 + 3² + 3³ + ... +  3¹¹

^b) chứng minh A chia hết cho 40

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

Dễ thấy \(B\)chia hết cho \(3\)do là tổng của các số hạng chia hết cho \(3\).

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

\(=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{119}+3^{120}\right)\)

\(=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{119}\left(1+3\right)\)

\(=4\left(3+3^3+...+3^{119}\right)⋮4\)

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

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

\(=3\left(1+3+3^2\right)+...+3^{118}\left(1+3+3^2\right)\)

\(=13\left(3+...+3^{118}\right)⋮13\)

a) \(B\)là tổng các số hạng chia hết cho \(3\)nên chia hết cho \(3\).

b) \(B=3+3^2+...+3^{120}\)

\(=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{119}+3^{120}\right)\)

\(=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{119}\left(1+3\right)\)

\(=4\left(3+3^3+...+3^{119}\right)⋮4\)

c) \(B=3+3^2+...+3^{120}\)

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

\(=3\left(1+3+3^2\right)+...+3^{118}\left(1+3+3^2\right)\)

\(=13\left(3+3^4+...+3^{118}\right)⋮13\)

C=(1+3+3^2)+3^3(1+3+3^2)+...+3^15(1+3+3^2)

=13(1+3^3+...+3^15) chia hết cho 13

\(A=3+3^2+3^3+...+3^{99}\\ \Rightarrow A=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{97}+3^{98}+3^{99}\right)\\ \Rightarrow A=3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+...+3^{97}\left(1+3+3^2\right)\\ \Rightarrow A=\left(1+3+3^2\right)\left(3+3^4+...+3^{97}\right)\\ \Rightarrow A=13\left(3+3^4+...+3^{97}\right)⋮13\)

\(A=3+3^2+3^3+...+3^{99}\\ 3A-A=3^{99}-1\\ A=\dfrac{3^{99}-1}{2}\)