có A=2(1+2)+2^3(1+2)+..+2^2019(1+2)

       =2.3+2^3.3+...+2^2019.3

       =3(2+2^3+...+2^2019)⋮3

Có A=2(1+2+2^2+2^3+2^4)+...+2^2016(1+2+2^2+2^3+2^4)

          =2.31+.....+2^2016.31

         =31.(2+....+2^2016)⋮31

`#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.`

a=2+2^2+2^3+...+2^10

a=(2+2^2)+(2^3+2^4)+...+(2^9+2^10)

a=2.(1+2)+2^3.(1+2)+...+2^9.(1+2)

a=3.(2+2^3+...+2^9)

=> a chia hết cho 3

a=2+2^2+2^3+...+2^10

a=(2+2^2+2^3+2^4+2^5)+(2^6+2^7+2^8+2^9+2^10)

a=2.(1+2+4+8+16)+2^6.(1+2+4+8+16)

a=31.(2+2^6)

=> a chia hết cho 31

chúc bạn học tốt nha

Cảm ơn bạn nhiều nha

^ nghĩa là gì

a, A = 2+2²+2³+...+2¹⁰

A = (2+2²)+(2³+2⁴)+...+(2⁹+2¹⁰)

A = 2(1+2) + 2³(1+2) +.....+ 2⁹(1+2)

A = 2.3 + 2³.3 +....+ 2⁹.3

A = 3.(2+2³+...+2⁹) chia hết cho 3 (đpcm)

b, A = 2+2²+2³+...+2¹⁰

A = (2+2²+2³+2⁴+2⁵)+(2⁶+2⁷+2⁸+2⁹+2¹⁰)

A = 2(1+2+2²+2³+2⁴) + 2⁶.(1+2+2²+2³+2⁴)

A = 2.31 + 2⁶.31

A = 31.(2+2⁶) chia hết cho 31 (Đpcm)

Bài 3:

\(A=5+5^2+..+5^{12}\)

\(5A=5\cdot\left(5+5^2+..5^{12}\right)\)

\(5A=5^2+5^3+...+5^{13}\)

\(5A-A=\left(5^2+5^3+...+5^{13}\right)-\left(5+5^2+...+5^{12}\right)\)

\(4A=5^2+5^3+...+5^{13}-5-5^2-...-5^{12}\)

\(4A=5^{13}-5\)

\(A=\dfrac{5^{13}-5}{4}\)

A=7 mu 2020 mu 2019-3 mu 2016 mu 2015 :5 chung to A la so chan