Lời giải:
$A=(2+2^2)+(2^3+2^4)+....+(2^{99}+2^{100})$
$=2(1+2)+2^3(1+2)+...+2^{99}(1+2)$

$=2.3+2^3.3+...+2^{99}.3$

$=3(2+2^3+...+2^{99})\vdots 3$

Ta có đpcm.

*Sửa lại đề*

A = 2¹+ 2²+ 2³+ 2⁴ + .. + 2¹⁰⁰

A = (2¹+2²) + (2³+ 2⁴) +...+ (2⁹⁹+ 2¹⁰⁰)

A = 2.(1+2) + 2³.(1+2) + .. + 2⁹⁹. (1+2)

A = 2.3 + 2³. 3 + .. + 2⁹⁹.3

A = 3 . (2¹ + 2³ + .... + 2⁹⁹)

Mà 3 chia hết cho 3 

=> A chia hết cho 3

\(A=2+2^2+2^3+2^4+...+2^{100}\)

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

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

\(=6\left(1+2^2+...+2^{98}\right)\)chia hết cho \(6\).

\(A+2+2^2+2^3+...+2^{100}\)

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

\(=6+2^2.6+...+2^{98}.6=6\left(1+2^2+...+2^{98}\right)⋮6\)

\(A=2+2^2+2^3+2^4+...+2^{100}\)

\(=2\cdot3+...+2^{99}\cdot3\)

\(=6\left(1+...+2^{99}\right)⋮6\)

Cách làm nữa ạ.CẢm ơn

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

\(=6+2^2.6+...+2^{98}.6=6\left(1+2^2+...+2^{98}\right)⋮6\)

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

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

\(=6+6.2^2+...+6.2^{98}\)

\(=6\left(1+2^2+...+2^{98}\right)⋮6\)

\(A=2+2^2+2^3+2^4+...+2^{100}\)

\(=2\cdot3+2^3\cdot3+...+2^{99}\cdot3\)

\(=6\left(1+2^2+...+2^{98}\right)⋮6\)

1: \(A=2+2^2+2^3+2^4+...+2^{97}+2^{98}+2^{99}+2^{100}\)

\(=2\left(1+2+2^2+2^3\right)+...+2^{97}\left(1+2+2^2+2^3\right)\)

\(=15\left(2+2^5+...+2^{97}\right)\)

\(=30\left(1+2^4+...+2^{96}\right)⋮30\)

2:

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

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

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

\(=12\left(1+3^2+...+3^{2020}\right)⋮12\)