\(A=\left(3+3^2+3^3\right)+...+\left(3^{58}+3^{59}+3^{60}\right)\\ A=3\left(1+3+3^2\right)+...+3^{58}\left(1+3+3^2\right)\\ A=\left(1+3+3^2\right)\left(3+...+3^{58}\right)\\ A=13\left(3+...+3^{58}\right)⋮13\)

\(M=\left(2+2^2+2^3+2^4\right)+...+\left(2^{17}+2^{18}+2^{19}+2^{20}\right)\\ M=\left(2+2^2+2^3+2^4\right)+...+2^{16}\left(2+2^2+2^3+2^4\right)\\ M=\left(2+2^2+2^3+2^4\right)\left(1+...+2^{16}\right)\\ M=30\left(1+...+2^{16}\right)⋮5\)

Câu 1: 

$A=(2+2^2)+(2^3+2^4)+(2^5+2^6)+....+(2^{2019}+2^{2020})$

$=2(1+2)+2^3(1+2)+2^5(1+2)+....+2^{2019}(1+2)$

$=(1+2)(2+2^3+2^5+...+2^{2019})=3(2+2^3+2^5+...+2^{2019})\vdots 3$

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$A=2+(2^2+2^3+2^4)+(2^5+2^6+2^7)+....+(2^{2018}+2^{2019}+2^{2020})$

$=2+2^2(1+2+2^2)+2^5(1+2+2^2)+....+2^{2018}(1+2+2^2)$

$=2+(1+2+2^2)(2^2+2^5+....+2^{2018})$

$=2+7(2^2+2^5+...+2^{2018})$

$\Rightarrow A$ chia $7$ dư $2$.

Câu 2:

$B=(3+3^2)+(3^3+3^4)+....+(3^{2021}+3^{2022})$
$=3(1+3)+3^3(1+3)+...+3^{2021}(1+3)$

$=(1+3)(3+3^3+...+3^{2021})=4(3+3^3+....+3^{2021})\vdots 4$

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$B=(3+3^2+3^3)+(3^4+3^5+3^6)+...+(3^{2020}+3^{2021}+3^{2022})$

$=3(1+3+3^2)+3^4(1+3+3^2)+....+3^{2020}(1+3+3^2)$

$=(1+3+3^2)(3+3^4+...+3^{2020})=13(3+3^4+...+3^{2020})\vdots 13$ (đpcm)

A=\((1+2)+\left(2^2+2^3\right)+...+\left(2^{19}+2^{20}\right)\)

A=\(3.1+2^2\left(1+2\right)+...+2^{19}\left(1+2\right)\)

A=\(3.1+3.2^2+...+3.2^{19}\)

A=\(3\left(1+2^2+...+2^{19}\right)\)\(⋮3\)

Vậy A\(⋮3\)

A=(1+2)+(22+23)+...+(219+220)(1+2)+(22+23)+...+(219+220)

A=3.1+22(1+2)+...+219(1+2)3.1+22(1+2)+...+219(1+2)

A=3.1+3.22+...+3.2193.1+3.22+...+3.219

A=3(1+22+...+219)3(1+22+...+219)⋮3⋮3

NÊN  A⋮3

A = 8⁸ + 2²⁰

= (2³)⁸ + 2²⁰

= 2²⁴ + 2²⁰

= 2²⁰.(2⁴ + 1)

= 2²⁰.17 ⋮ 17

Vậy A ⋮ 17

Đề sai nghe

kết hợp theo công thức thì số kết thúc phải là 2¹⁹ hoặc là 2²¹  mới kết hợp được
Đừng có đánh giá người khác như thế chứ ;-;

A = 2 + 22 + 23 + 24 + ... + 219 + 220

A = (2 + 22) + (23 + 24) +... + (219 + 220)

A = 2.(1+2) + 23.(1 + 2) +... + 219.(l + 2)

A = 2.3 + 23.3 +...+ 219.3 Do đó A chia hết cho 3

do đó A chia hết cho 3