编写程序代码
Private Sub Form_Click()
Dim n As Integer
Dim s As Single
n = 1: s = 0
Do While n <= 180
s = s + n / (1 + n * n)
n = n + 1
Loop
s = 0.1 * Int(10 * s + 0.5) 'Int函数对数据进行四舍五入
Print "s="; s
End Sub
计算并输出下列级数和s=1+1\/(1+2)+1\/(1+2+3)+...+1\/(1+2+3+...+n)
include <stdio.h> int main(void){ int i,n;double s=0,s0=0;printf("请出入n的值:");scanf("%d",&n);for(i=1;i<=n;i++){ s0+=i;\/\/先计算分母 s+=1.0\/s0;\/\/再由s0推算出s } printf("s=%lf\\n",s);return 0;} 验证：n=1 n=2 n=3 嗯，写错了个字，是“请...
1+1\/(1+2)+1\/(1+2+3)+…+1\/(1+2+3+…100)=
1+2=2*3\/2 1+2+3=3*4\/2 1+2+3+4=4*5\/2 1+2+3+……+100=100*101\/2 所以，1+1\/(1+2)+1\/(1+2+3)+1\/(1+2+3+4)+...+1\/(1+2+3+...+2006)=1+2\/（2*3）+2\/（3*4）+2\/（4*5）+……+2\/（100*101）=2[（1\/2+1\/（2*3）+1\/（3*4）+1\/（4*5...
...s=1+(1+2)+(1+2+3)+(1+2+3+4)+… +(1+2+3+… +n)。用Python来一下...
print("s=")print(sum(lst3))这是最简单的 调试结果
1+1\/(1+2)+1\/(1+2+3)+1\/(1+2+3+4)+...1\/(1+2+3+...99+100)
= 1 + 2*(1\/2 - 1\/101)= 2 - 2\/101 = 200\/101
...s=1+1\/(1+2)+1\/(1+2+3)+1\/(1+2+3+4)+...+1\/(1+2+3+...+n)_百度知...
include <stdio.h> int main(){ double s=0;int n,t=0;printf("请输入n\\n");scanf("%d",&n);int i;for(i=1;i<=n;i++){ t+=i;s+=1.0\/t;} printf("结果为：%f",s);return 0;}
...求S=1\/(1*2)+1\/(2*3)+1\/(3*4)+……前50项之和。
include<stdio.h> int main(){int i;float y=0;for(i=1;i<=50;i+=2)y+=1.0\/(i*(i+1));printf("%g\\n",y);return 0;}
计算巧算1+1\/(1+2)+1\/(1+2+3)+1\/(1+2+3+4)+……1\/(1+2+3+……+100...
1+2=2*3\/2 1+2+3=3*4\/2 1+2+3+4=4*5\/2 1+2+3+……+100=100*101\/2 所以，1+1\/(1+2)+1\/(1+2+3)+1\/(1+2+3+4)+...+1\/(1+2+3+...+100)=1+2\/（2*3）+2\/（3*4）+2\/（4*5）+……+2\/（100*101）=2[（1\/2+1\/（2*3）+1\/（3*4）+1\/（4*5...
怎么用C语言计算S=1+1\/(1+2)+1\/(1+2+3)+……+1\/(1+2+3+……+100)
include "stdio.h"int main(int argv,char *argc[]){double s;int i,t;for(s=t=0,i=1;i<101;s+=1.0\/(t+=i),i++);printf("The result are %f\\n",s);return 0;}运行结果：
求极限Xn=1\/(n^2+1)+2\/(n^2+2)+3\/(n^2+3)+……+n\/(n^2+n)
简单计算一下即可，答案如图所示
初一数学题1+1\/(1+2)+1\/(1+2+3)+1\/(1+2+3+4)+...+1\/(1+2+3+...+100...
因为：1\/(1+2)=1\/3=2*(1\/2-1\/3)1\/(1+2+3)=1\/6=2*(1\/3-1\/4)1\/(1+2+3+4)=1\/10=2*(1\/4-1\/5)...同理 1\/(1+2+3+...+100)=1\/5050=2*（1\/100-1\/101)所以 1+1\/(1+2)+1\/(1+2+3)+1\/(1+2+3+4)+...+1\/(1+2+3+...+100)=1+2*【(1\/2-1...
