`#3107.101107`

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

\(= (2 + 2^2) + (2^3 + 2^4) + ... + (2^{2021} + 2^{2022})\)

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

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

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

Vì \(3(2 + 2^3 + ... + 2^{2021})\) \(\vdots\) \(3\)

`\Rightarrow A \vdots 3`

Vậy, `A \vdots 3.`

2021²⁰²⁰ tận cùng là 1 ; 2025²⁰²⁵ tận cùng là 5

2022¹⁰ = (2022⁴)².2022² = (...6)².(...4) = (...6).(...4) tận cùng là 4 (vì 6.4 = 24 tận cùng là 4)

  S= 5+5²+5³+...+5²⁰²⁰+5²⁰²¹

 5S=5²+5³+5⁴+...+5²⁰²¹+5²⁰²²

 5S - S=4S=5²⁰²²-5

  Ta có: 4S+5=5²⁰²²

             =4S -5 +5 =5²⁰²²

              => 4S=5²⁰²²

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 = (5+5^2+5^3)+(5^4+5^5+5^6)+...+(5^2020+5^2021+5^2022)

= 5(1+5+5^2)+5^4(1+5+5^2)+...+5^2020(1+5+5^2)

= 5.31+5^4.31+...+5^2020.31

= 31(5+5^4+...+5^2020) chia hết cho 31

\(C=4+4^2+4^3+...+4^{2021}+4^{2022}\)

\(=\left(4+4^2\right)+\left(4^3+4^4\right)+...+\left(4^{2021}+4^{2022}\right)\)

\(=4.\left(1+4\right)+4^3.\left(1+4\right)+...+4^{2021}.\left(1+4\right)\)

\(=4.5+4^3.5+...+4^{2021}.5\)

\(=5.\left(4+4^3+...+4^{2021}\right)⋮5\)

Vậy \(C⋮5\)

a, \(S=3^0+3^2+3^4+3^6+...+3^{2020}\)

\(\Leftrightarrow3^2S=3^2+3^4+3^6+3^8+...+3^{2022}\)

\(\Leftrightarrow3^2S-S=3^{2022}-3^0\)

\(\Leftrightarrow9S-S=3^{2022}-1\)

\(\Leftrightarrow8S=3^{2022}-1\Leftrightarrow S=\frac{3^{2022}-1}{8}\)

b,\(S=3^0+3^2+3^4+3^6+...+3^{2020}\)

\(=\left(3^0+3^2+3^4\right)+\left(3^6+3^8+3^{10}\right)+...+\left(3^{2016}+3^{2018}+3^{2020}\right)\)

\(=\left(1+3^2+3^4\right)+3^6\left(1+3^2+3^4\right)+...+3^{2016}\left(1+3^2+3^4\right)\)

\(=\left(1+3^2+3^4\right)\left(1+3^6+...+3^{2016}\right)\)

\(=91\left(1+3^6+...+3^{2016}\right)=13.7\left(1+3^6+...+3^{2016}\right)⋮7\)

=> đpcm

Tham khảo :

a, $S=3^0+3^2+3^4+3^6+...+3^{2020}$

$\iff 3^{2}S=3^2+3^4+3^6+3^8+...+3^{2022}$

$\iff 3^{2}S-S=3^{2022}-3^0$

$\iff 9S-S=3^{2022}-1$

$\iff 8S=3^{2022}-1\iff S=\frac{3^{2022}-1}{8}$

b,$S=3^0+3^2+3^4+3^6+...+3^{2020}$

$=\left(3^0+3^2+3^4\right)+\left(3^6+3^8+3^{10}\right)+...+\left(3^{2016}+3^{2018}+3^{2020}\right)$

$=\left(1+3^2+3^4\right)+3^6\left(1+3^2+3^4\right)+...+3^{2016}\left(1+3^2+3^4\right)$

$=\left(1+3^2+3^4\right)\left(1+3^6+...+3^{2016}\right)$

$=91\left(1+3^6+...+3^{2016}\right)=13.7\left(1+3^6+...+3^{2016}\right)⋮7$

=> (đpcm)

A<1/1*2+1/2*3+...+1/2021*2022

=>A<1-1/2+1/2-1/3+...+1/2021-1/2022<1