参考解法：

 #!/usr/bin/python # -*- coding: UTF-8 -*- s=0 def fact(n): if n==1: return 1 return n*fact(n-1) for n in range(1,21): a=fact(n) s+=a print(s) 

参考解法：

 #!/usr/bin/python # -*- coding: UTF-8 -*- sum1=0 for each in range(1,21): sum2=1 while each>1: sum2=sum2*each each-=1 sum1+=sum2 print(sum1) 

参考方法：

 #!/usr/bin/python3 n=int(input("Enter a number:")) s=0 def cal(n): pro=1 for i in range(1,n+1): pro=pro*i return pro for i in range(1,n+1): if i!=n: print("%d! + "%i,end='') else: print("%d! = "%i,end='') s=s+cal(i) print(s) 

使用两个递归函数实现，其中jie()生成阶乘，sum()对阶乘求和

 #!/usr/bin/python # -*- coding: UTF-8 -*- def jie(n): if n==1: return 1 else: return n*jie(n-1) def sum(n): if n==1: return jie(1) else: return jie(n)+sum(n-1) #a=int(input("请输入一个正整数：\n")) #print(sum(a)) print(sum(20)) 

Python3下的实现

#python 3 from functools import reduce a = 0 for n in range(1, 21): a += reduce(lambda x, y: x * y, range(1, n + 1)) print(a)