Prove that for all whole values of n the value of the expression:
n(n–1)–(n+3)(n+2) is divisible by 6.

Expand:
[tex]n(n-1)-(n+3)(n+2)=(n^2-n)-(n^2+5n+6)=-6n-6[/tex]
Then we can write
[tex]n(n-1)-(n+3)(n+2)=6\boxed{(-n-1)}[/tex]
which means [tex]6\mid n(n-1)-(n+3)(n+2)[/tex] as required.