(n+2)! - (3n+2) = (n+2)(n+1)! - (3n+2)
= (n+2)(n+1)! - 3(n+2)
= (n+2)[(n+1)! - 3]
= (n+2)(n! + n - 3)
= n!n + n! + n^2 + 2n - 3n - 6
= n!n + n! + n^2 - n - 6

Ответ: n!n + n! + n^2 - n - 6