\(A=1+3+3^2+..........+3^{11}\)
\(\Leftrightarrow A=\left(1+3\right)+\left(3^2+3^3\right)+.........+\left(3^{10}+3^{11}\right)\)
\(\Leftrightarrow A=1\left(1+3\right)+3^2\left(1+3\right)+.........+3^{10}\left(1+3\right)\)
\(\Leftrightarrow A=1.4+3^2.4+.......+3^{10}.4\)
\(\Leftrightarrow A=4\left(1+3^2+..........+3^{10}\right)⋮4\left(đpcm\right)\)
A = 1 + 3 + 32 + 33 + ... + 311
A = ( 1 + 3 ) + ( 32 + 33 ) + ... + ( 310 + 311 )
A = 4 + 32 . ( 1 + 3 ) + ... + 310 . ( 1 + 3 )
A = 4 + 32 . 4 + ... + 310 . 4
A = 4 . ( 1 + 32 + ... + 310 ) \(⋮\) 4 ( Vì trong tích có một thừa số chia hết cho 4 )
~ Chúc bạn học giỏi ! ~
A = 1 + 3 + 32 + 33 + ... + 311
A = ( 1 + 3 + 32 ) + ... + ( 39 + 310 + 311 )
A = 13 + ... + 39 . ( 1 + 3 + 32 )
A = 13 + ... + 39 . 13
A = 13 . ( 1 + ... + 39 ) \(⋮\) 13 ( Vì trong tích có một thừa số chia hết cho 13 )
~ Chúc bạn học giỏi ! ~
a) Ta có : \(A=1+3+3^2+3^3+...+3^{11}\)
\(=\left(1+3\right)+\left(3^2+3^3\right)+\left(3^4+3^5\right)+...+\left(3^{10}+3^{11}\right)\)
\(=4+3^2\left(3+1\right)+3^4\left(3+1\right)+...+3^{10}\left(3+1\right)\)
\(=4.\left(1+3^2+3^4+...+3^{10}\right)⋮4\)
\(\Rightarrow A⋮4\)
\(\Rightarrowđpcm\)
b) Ta có : \(A=1+3+3^2+3^3+...+3^{11}\)
\(=\left(1+3^2\right)+\left(3+3^3\right)+...+\left(3^9+3^{11}\right)\)
\(=10+3\left(1+3^2\right)+3^4\left(1+3^2\right)+...+3^9\left(1+3^2\right)\)
\(=10+3.10+3^4.10+...+3^9.10\)
\(=10.\left(1+3+3^4+...+3^9\right)⋮10\) \(\Rightarrow A⋮10\) \(\Rightarrowđpcm\) c) Ta có : \(A=1+3+3^2+3^3+...+3^{11}\) \(=\left(1+3+3^2\right)+...+\left(3^9+3^{10}+3^{11}\right)\) \(=13+3^3.\left(1+3+3^2\right)+...+3^9.\left(1+3+3^2\right)\) \(=13+3^3.13+...+3^9.13\) \(=13.\left(1+3^3+3^6+3^9\right)⋮13\) \(\Rightarrow A⋮13\) \(\Rightarrowđpcm\)
