a,

`3A=3+3^3+3^3+...+3^{53}`

`3A-A=(3+3^3+3^3+...+3^{53})-(1+3+3^3+3^3+...+3^{52})`

`2A=3^{53}-1`

`A=(3^{53}-1)/2`

b,

`A=1+3+3^3+3^3+...+3^{52}`

`A=(1+3+3^2)+(3^3+3^4+3^5)+....+(3^{50}+3^{51}+3^{52})`

`A=(1+3+3^2)+3^3*(1+3+3^2)+....+3^{50}*(1+3+3^2)`

`A=(1+3+3^2)*(1+3^3+....+3^{50})`

`A=13*(1+3^3+....+3^{50})`

Do `13 \vdots 13 => A=13*(1+3^3+....+3^{50})\vdots 13 `

Vậy `A \vdots 13 `

\(B=3+3^2+3^3+...+3^{100}\)

\(=3+\left(3^2+3^3+3^4\right)+...+\left(3^{98}+3^{99}+3^{100}\right)\)

\(=3+3^2\left(1+3+3^2\right)+...+3^{98}\left(1+3+3^2\right)\)

\(=3+3^2.13+...+3^{98}.13\)

\(=3+13\left(3^2+...+3^{98}\right)\)

\(\Rightarrow B⋮̸13\)

\(\Rightarrow B:13\) dư 3.

Các bạn giải nhanh giúp mình nhé. Mình cần gấp. Thanks!

Lồn bâm

Gâu gâu 

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

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

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

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

=>A chia 13 dư 1

Bạn ko biết gõ số mũ à gõ thế này bố ai mà hiểu được

Lời giải:
$A=1+(3+3^2+3^3)+(3^4+3^5+3^6)+....+(3^{2014}+3^{2015}+3^{2016})$

$=1+3(1+3+3^2)+3^4(1+3+3^2)+...+3^{2014}(1+3+3^2)$
$=1+3.13+3^4.13+....+3^{2014}.13$

$=1+13(3+3^4+...+3^{2014})$ 

$\Rightarrow A-1\vdots 13(1)$

Mặt khác:
$A=1+(3+3^2+3^3+3^4)+....+(3^{2013}+3^{2014}+3^{2015}+3^{2016})$
$=1+3(1+3+3^2+3^3)+....+3^{2013}(1+3+3^2+3^3)$
$=1+(3+...+3^{2013})(1+3+3^2+3^3)$

$=1+40(3+....+3^{2013})$

$\Rightarrow A-1\vdots 5(2)$

Từ $(1); (2)$ mà $(5,13)=1$ nên $A-1\vdots (5.13)$ hay $A-1\vdots 65$

$\Rightarrow A$ chia $65$ dư $1$

B=3+3²+3³+..... +3¹00 

B=3²+3³+3⁴+... 3¹00+3

B=3²(1+3+3²) +... +3 98(1+3+3²) +3

B=3²•13+... +3 98•13+3

=) 3²•13+3 98•13 chia hết cho 13

=) Số dư là 3