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

\(=13\left(1+...+3^7\right)⋮13\)

A = 8⁸ + 2²⁰

= (2³)⁸ + 2²⁰

= 2²⁴ + 2²⁰

= 2²⁰.(2⁴ + 1)

= 2²⁰.17 ⋮ 17

Vậy A ⋮ 17

b) \(A=3+3^2+3^3+...+3^{60}\)

\(A=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{58}+3^{59}+3^{60}\right)\)

\(A=3\left(1+3+3^2\right)+3^4\cdot\left(1+3+3^2\right)+...+3^{58}\cdot\left(1+3+3^2\right)\)

\(A=3\cdot13+3^4\cdot13+...+3^{58}\cdot13\)

\(A=13\cdot\left(3+3^4+...+3^{58}\right)\)

Vậy A chia hết cho 13

a) \(A=3+3^2+...+3^{60}\)

\(A=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{59}+3^{60}\right)\)

\(A=3\cdot\left(1+3\right)+3^3\cdot\left(1+3\right)+...+3^{59}\cdot\left(1+3\right)\)

\(A=4\cdot\left(3+3^3+...+3^{59}\right)\)

Nên A chia hết cho 4

S = 3 + 3 2 + 3 3 + 3 4 + 3 5 + 3 6 + 3 7 + 3 8 + 3 9      =   3 + 3 2 + 3 3 + 3 4 + 3 5 + 3 6 + 3 7 + 3 8 + 3 9      = 39 + 3 3   .   39   +   3 6   .   39      = 39 . 1 + 3 3 + 3 6   ⋮   − 39  

Vậy S chia hết cho -39

S = 3 + 3 2 + 3 3 + 3 4 + 3 5 + 3 6 + 3 7 + 3 8 + 3 9      =   3 + 3 2 + 3 3 + 3 4 + 3 5 + 3 6 + 3 7 + 3 8 + 3 9      = 39 + 3 3   .   39   +   3 6   .   39      = 39. 1 + 3 3 + 3 6   ⋮   − 39  

Vậy S chia hết cho -39

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

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

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

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

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

Vậy.... 

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

Số các số hạng là: 101 – 0 + 1 = 102 số.
Ta nhận thấy:
1 + 3 + 32 = 1 + 3 + 9 = 13;
33 + 34 + 35 = 33(1 + 3 + 32) = 33.13;
…
Mà 102 có tổng các chữ số là 1 + 0 + 2 = 3 chia hết cho 3 nên 102 chia hết cho 3, nghĩa là:
A = (1 + 3 + 32) + (33 + 34 + 35) + … + (399 + 3100 + 3101)
= (1 + 3 + 32) + 33(1 + 3 + 32) + … + 399(1 + 3 + 32)
= 13 + 33.13 + … + 399.13
= 13.(1 + 33 + … + 399) chia hết cho 13.
Vậy A chia hết cho 13.

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

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

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

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