Question

CMR :

\(n^3+5n⋮6\forall n\in Z\)

Answer 1

ta có : 

\(n^3+5n=n^2-n+6n\)

                \(=\left(n-1\right)n\left(n+1\right)+6n\)

mà \(\left(n-1\right)n\left(n+1\right)⋮2;3\)

\(\Rightarrow\left(n-1\right)n.\left(n+1\right)⋮6\)

\(\Rightarrow6n⋮6\)

\(\Rightarrow n^3+5n⋮6\)

Answer 2

sorry mk nhầm ! 

chỗ : \(n^2-n+6n\)phải thành 

\(n^3-n+6n\)

Answer 3

\(n^3+5n=n.\left(n^2+5\right)=n.\left(n^2-1\right)+6n=n.\left(n-1\right).\left(n+1\right)+6n\)

\(\hept{\begin{cases}n.\left(n+1\right).\left(n-1\right)⋮2,3\text{ hay }n.\left(n+1\right).\left(n-1\right)⋮6\forall n\in Z\\6n⋮6\forall n\in Z\end{cases}}\)

=> đpcm

p/s: cách này có vẻ gọn hơn  =)