A=[1+3+3^2+3^3]+...+[3^2018+3^2019+3^2020+3^2021]

A=1 nhân[1+3+3^2+3^3]+...+3^2018 nhân [1+3+3^2+3^3]

A=[1+3+3^2+3^3] NHÂN[1+...+3^2018

A=40 nhân [1+...+3^2018]

=> A chia hết cho 40

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

\(M=1+3+3^2+............+3^{100}\)

\(\Leftrightarrow M=1+3+\left(3^2+3^3+3^4\right)+\left(3^5+3^6+3^7\right)+.......+\left(3^{98}+3^{99}+3^{100}\right)\)

\(\Leftrightarrow M=4+3^2\left(1+3+3^2\right)+3^5\left(1+3+3^2\right)+......+3^{98}\left(1+3+3^2\right)\)

\(\Leftrightarrow M=4+3^2.13+3^5.13+.........+3^{98}.13\)

\(\Leftrightarrow M=4+13\left(3^2+3^5+..........+3^{98}\right)\)

Mà \(13\left(3^2+3^5+......+3^{98}\right)⋮13\)

\(4:13\left(dư4\right)\)

\(\Leftrightarrow M:13\left(dư4\right)\)

b, tương tự

\(A=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{89}+3^{90}\right)\\ A=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{89}\left(1+3\right)\\ A=3\cdot4+3^3\cdot4+...+3^{89}\cdot4\\ A=4\left(3+3^3+...+3^{89}\right)⋮4\)

A = ( 3 + 3 2 ) + ( 3 3 + 3 4 ) + . . . + ( 3 89 + 3 90 )

A = 3 ( 1 + 3 ) + 3 3 ( 1 + 3 ) + . . . + 3 89 ( 1 + 3 )

A = 3 ⋅ 4 + 3 3 ⋅ 4 + . . . + 3 89 ⋅ 4

A = 4 ( 3 + 3 3 + . . . + 3 89 ) ⋮ 4

Lồn bâm

Gâu gâu 

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

\(\Rightarrow A=\left(3+3^2+3^3+3^4\right)+.......+\left(3^{97}+3^{98}+3^{99}+3^{100}\right)\)

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

\(\Rightarrow A=3.40+.........+3^{97}.40\)

\(\Rightarrow A=40.\left(3+.......+3^{97}\right)\)

\(\Rightarrow A⋮40\)( 1 )

Vì \(A\)là tổng của các bậc lũy thừa của 3 nên \(A⋮3\)( 2 )

Từ ( 1 ) và ( 2 ) suy ra : \(A⋮40.3\)

\(\Rightarrow A⋮120\)

Vậy \(A⋮120\)( ĐPCM )