Dả sử n\(⋮3\Leftrightarrow n=3k\)
Thay vào, ta có :
\(3k\left(3k+1\right)\left(3k+5\right)⋮3\) ( đúng )
Còn nếu \(n⋮̸3\Leftrightarrow n=3k+1;n=3k+2\)
Khi \(n=3k+1,\)ta có :
\(\left(3k+1\right)\left(3k+2\right)\left(3k+6\right)=3\left(3k+1\right)\left(3k+2\right)\left(k+1\right)⋮3\) ( đúng )
Khi \(n=3k+2,\) ta có :
\(\left(3k+2\right)\left(3k+3\right)\left(3k+7\right)=3\left(3k+2\right)\left(k+1\right)\left(3k+7\right)⋮3\) ( đúng )
Vậy \(n\left(n+1\right)\left(n+3\right)⋮3\forall n\)