∵f1（x）= 1?x ex = (?1)1(x?1) ex ，f2（x）= x?2 ex = (?1)2(x?2) ex ，f3（x）= 3?x ex = (?1)3(x?3) ex ，…，由此归纳可得：fn（x）= (?1)n(x?n) ex ，故答案为： (?1)n(x?n) ex
