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

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

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

\(A=2+2^2+2^3+...+2^{2020}+2^{2021}+2^{2022}\\=(2+2^2)+(2^3+2^4)+(2^5+2^6)+...+(2^{2021}+2^{2022})\\=2\cdot(1+2)+2^3\cdot(1+2)+2^5\cdot(1+2)+...+2^{2021}\cdot(1+2)\\=2\cdot3+2^3\cdot3+2^5\cdot3+...+2^{2021}\cdot3\\=3\cdot(2+2^3+2^5+..+2^{2021})\)

Vì \(3\cdot\left(2+2^3+2^5+...+2^{2021}\right)⋮3\)

nên \(A⋮3\).

\(Toru\)

A=(2+2²)+2²(2+2²)+...+2²⁰²⁰(2+2²)

A= 6.1+2².6+...+2²⁰²⁰.6

A=6(1+2²+...+2²⁰²⁰) chia hết cho 3

vậy A chia hết cho 3

Câu 1: 

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

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

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

-----------------

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

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

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

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

$\Rightarrow A$ chia $7$ dư $2$.

Câu 2:

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

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

-------------------

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

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

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

xam xi

Lời giải:
$S=(2+2^2)+(2^3+2^4)+....+(2^{23}+2^{24})$

$=2(1+2)+2^3(1+2)+....+2^{23}(1+2)$

$=(1+2)(2+2^3+...+2^{23})$

$=3(2+2^3+...+2^{23})\vdots 3$

b.

$S=2+2^2+2^3+...+2^{23}+2^{24}$

$2S=2^2+2^3+2^4+....+2^{24}+2^{25}$

$\Rightarrow 2S-S=2^{25}-2$

$\Rightarrow S=2^{25}-2$

Ta có:

$2^{10}=1024=10k+4$

$\Rightarrow 2^{25}-2=2^5.2^{20}-2=32(10k+4)^2-2=32(100k^2+80k+16)-2$
$=10(320k^2+8k+51)\vdots 10$

$\Rightarrow S$ tận cùng là $0$