1: \(A=2+2^2+2^3+2^4+...+2^{97}+2^{98}+2^{99}+2^{100}\)

\(=2\left(1+2+2^2+2^3\right)+...+2^{97}\left(1+2+2^2+2^3\right)\)

\(=15\left(2+2^5+...+2^{97}\right)\)

\(=30\left(1+2^4+...+2^{96}\right)⋮30\)

2:

\(B=3+3^2+3^3+...+3^{2022}\)

\(=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2021}+3^{2022}\right)\)

\(=\left(3+3^2\right)+3^2\left(3+3^2\right)+...+3^{2020}\left(3+3^2\right)\)

\(=12\left(1+3^2+...+3^{2020}\right)⋮12\)

\(3+3^2+...+3^{2022}\)

\(=\left(3+3^2+3^3\right)+...+\left(3^{2020}+3^{2021}+3^{2022}\right)\)

\(=3\cdot\left(1+3+9\right)+3^4\cdot\left(1+3+9\right)+...+3^{2020}\cdot\left(1+3+9\right)\)

\(=3\cdot13+3^4\cdot13+...+3^{2020}\cdot13\)

\(=13\cdot\left(3+3^4+...+3^{2020}\right)\) ⋮ 13 

Vậy.... 

Ta có: \(a=\frac13+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+\cdots+\frac{2022}{3^{2022}}\)

=>\(3a=1+\frac23+\frac{3}{3^2}+\frac{4}{3^3}+\cdots+\frac{2022}{3^{2021}}\)

=>\(3a-a=1+\frac23+\frac{3}{3^2}+\cdots+\frac{2022}{3^{2021}}-\frac13-\frac{2}{3^2}-\frac{3}{3^3}-\cdots-\frac{2022}{3^{2022}}\)

=>\(2a=1+\frac13+\frac{1}{3^2}+\cdots+\frac{1}{3^{2021}}-\frac{2022}{3^{2022}}\)

Đặt \(b=\frac13+\frac{1}{3^2}+\cdots+\frac{1}{3^{2021}}\)

=>\(3b=1+\frac13+\cdots+\frac{1}{3^{2020}}\)

=>\(3b-b=1+\frac13+\ldots+\frac{1}{3^{2020}}-\frac13-\frac{1}{3^2}-\cdots-\frac{1}{3^{2021}}\)

=>\(2b=1-\frac{1}{3^{2021}}=\frac{3^{2021}-1}{3^{2021}}\)

=>\(b=\frac{3^{2021}-1}{2\cdot3^{2021}}\)

Ta có: \(2a=1+\frac13+\frac{1}{3^2}+\cdots+\frac{1}{3^{2021}}-\frac{2022}{3^{2022}}\)

=>\(2a=1+\frac{3^{2021}-1}{2\cdot3^{2021}}-\frac{2022}{3^{2022}}=1+\frac{3^{2022}-3-4044}{2\cdot3^{2022}}=1+\frac12-\frac{4047}{2\cdot3^{2022}}\)

=>\(2a<\frac32\)

=>\(a<\frac34\)

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 \(A\vdots13\)

\(\text{#}Toru\)

\(A=1+3+3^2+3^3+...+3^{2022}\)

\(=1+\left(3+3^2+3^3\right)+...+\left(3^{2020}+3^{2021}+3^{2022}\right)\)

\(=1+3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+...+3^{2020}\left(1+3+3^2\right)\)

\(=1+13\left(3+3^4+...+3^{2020}\right)\)

=>A chia 13 dư 1

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2020 đâu có chia hết cho 3

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phải là chứng minh A chia hết cho 121