mikko biết

\(A=5+4^2+...+4^{2021}\\ A=4^0+4^1+...+4^{2021}\\ 4A=4^1+4^2+...+4^{2022}\\ 4A-A=\left(4^1+4^2+...+4^{2022}\right)-\left(4^0+4^1+...+4^{2021}\right)\\ 3A=4^{2022}-1\\ 3A+1=4^{2022}⋮4^{2021}\)

Lời giải:
$A-1=4+4^2+4^3+...+4^{2020}+4^{2021}$
$4(A-1)=4^2+4^3+4^4+....+4^{2021}+4^{2022}$

$\Rightarrow 4(A-1)-(A-1)=4^{2022}-4$

$3(A-1)=4^{2022}-4$

$\Rightarrow 3A+1=4^{2022}\vdots 4^{2021}$ 

Lg:

Ta có :A=5+4^2+4^3+...+4^2020+4^2021

4A=20+4^3+4^4+...+4^2021+4^2022

4A-A=(20+4^3+4^4+...+4^2021+4^2022)-(5+4^2+4^3+...+4^2020+4^2021)

3A=4^2022-4^2+20-5

3A=4^2022-16+15

3A+1=4^2022-16+15+1

3A+1=4^2022-16+16

3A+1=4^2022⋮4^2021

Vậy 3A+1⋮4^2021

Sửa đề: \(2021^3+2021^4+2021^5+2021^6\)

Ta có: \(2021^3+2021^4+2021^5+2021^6\)

\(=2021^3\left(1+2021\right)+2021^5\left(1+2021\right)\)

\(=2022\left(2021^3+2021^5\right)\) ⋮2022

\(A=1+2+2^2+2^3+...+2^{2021}\)

\(=7+2^3\cdot7+...+2^{2019}\cdot7\)

\(=7\left(1+...+2^{2019}\right)⋮7\)

\(7a+2b⋮2021;31a+9b⋮2021\)

\(\Rightarrow\hept{\begin{cases}9\left(7a+2b\right)-2\left(31a+9b\right)⋮2021\\31\left(7a+2b\right)-7\left(31a+9b\right)⋮2021\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}a⋮2021\\-b⋮2021\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}a⋮2021\\b⋮2021\end{cases}}\) (đpcm)

A=(1+3+3²)+(3³+3⁴+3⁵)+...+(3²⁰¹⁹+3²⁰²⁰+3²⁰²¹)                                                  A=(1+3+3²)+3³.(1+3+3²)+...+3²⁰¹⁹.(1+3+3²)

A=13+3³.13+...+3²⁰¹⁹.13

A=13.(1+3³+...+3²⁰¹⁹)chia hết cho 13

=>A  chia hết cho 13

A = (1+3+3 mũ 2)+(3 mũ 3+3 mũ 4+3 mũ 5)+....+(3 mũ 2019 + 3 mũ 2020 + 3 mũ 2021)

A = 1 (1 + 3 + 3 mũ 2) + 3 mũ 9 (1+3+3 mũ 3) +...+ 3 mũ 6057 ( 1+3+3 mũ 2)

A = 1.13 +3 mũ 9.13 + ... + 3 mũ 6057 . 13

A =13.(1+3 mũ 9 +...+ 3 mũ 6057)

13 chia hết cho 13 nên A chia hết cho 13

Chứng minh rằng: A = 3^2 + 3^3 + 3^4 + 3^5 + … + 3^2020 + 3^2021 chia hết cho 36 - Hoc24

\(A=\left(3^2+3^3\right)+3^2\left(3^2+3^3\right)+...+3^{2018}\left(3^2+3^3\right)\)

\(=36+3^2.36+...+3^{2018}.36=36\left(1+3^2+...+3^{2018}\right)⋮36\)