计算s=1+1/(1+2)+1/(1+2+3)+...+1/(1+2+3+...+n)python
时间: 2023-12-17 19:02:22 浏览: 89
以下是计算s的Python代码:
```python
n = int(input("请输入正整数n:"))
sum = 0
for i in range(1, n+1):
sum += 1 / sum(range(1, i+1)).sum()
print("sum = {:.10f}".format(sum))
```
阅读全文
相关推荐
















