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:

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

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

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

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

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

\(=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^{99}+3^{100}+3^{101}\right)\)

\(=\left(1+3+3^2\right)+3^3\left(1+3+3^2\right)+...+3^{99}\left(1+3+3^2\right)\)

\(=13\left(1+3^3+...+3^{99}\right)⋮13\)

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

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

a) B\(=\) 3 + 3² + 3³ + ... + 3⁶⁰ 

\(=\)(3+3²)+(3³+3⁴)+...+(3⁵⁹+3⁶⁰)

\(=\)3(1+3)+3³(1+3)+...+3⁵⁹(1+3)

\(=\)(3+1)(3+3³+...+3⁵⁹)

\(=\)4(3+3³+...+3⁵⁹)⋮4

⇒B⋮4

b) B\(=\)(3+3²+3³)+...+(3⁵⁸+3⁵⁹+3⁶⁰)

\(=\)3⁰(3+3²+3³)+...+3⁵⁷(3⁵⁸+3⁵⁹+3⁶⁰)

\(=\)3+3²+3³(3⁰+3³+3⁶+...+3⁵⁷)

\(=\)39(3⁰+3³+3⁶+...+3⁵⁷)⋮13

⇒ B⋮13

\(S=\left(1+3+3^2\right)+...+3^7\left(1+3+3^2\right)\)

\(=13\left(1+...+3^7\right)⋮13\)

Đặt A = 3² + 3³ + 3⁴ + ... + 3⁹⁹

= 3² + 3³ + (3⁴ + 3⁵ + 3⁶) + (3⁷ + 3⁸ + 3⁹) + ... + (3⁹⁷ + 3⁹⁸ + 3⁹⁹)

= 36 + 3⁴.(1 + 3 + 3²) + 3⁷.(1 + 3 + 3²) + ... + 3⁹⁷.(1 + 3 + 3²)

= 36 + 3⁴.13 + 3⁷.13 + ... + 3⁹⁷.13

= 36 + 13.(3⁴ + 3⁷ + ... + 3⁹⁷)

Do 36 không chia hết cho 13

13.(3⁴ + 3⁷ + ... + 3⁹⁷) ⋮ 13

⇒ 36 + 13.(3⁴ + 3⁷ + ... + 3⁹⁷) không chia hết cho 13

⇒ A không chia hết cho 13

Em xem lại đề nhé, có thể em viết thiếu số 3 rồi