C语言：计算并输出S=1+(1+2^0.5)+(1+2^0.5+3^0.5)...+(1+2^0.5+3^0.5+...+n^0.5)

//计算并输出S=1+(1+2^0.5)+(1+2^0.5+3^0.5)...+(1+2^0.5+3^0.5+...+n^0.5)

#include<math.h>
#include<stdio.h>
double fun(int n)
{
    double s=1,S=0;
    for (int i = 2; i <= n+1; i++)//这里判断条件为n+1.
    {
        S += s;
        s += sqrt(i);//2^0.5=2^(1/2)=根号下2=sqrt(2)
        printf("%f
", S);
        printf("%f
", s);
    }
    return S;
}
void main()
{
  FILE *wf;
  int n; 
  double s;
  printf("

Input n: ");
  scanf("%d",&n);
  s=fun(n);
  printf("

s=%f

",s);
/******************************/
  wf=fopen("out.dat","w");
  fprintf(wf,"%f",fun(20));
  fclose(wf);
/*****************************/
}

 //使用递归解决：

double f(int n)
{
    if (n == 1) return 1;
    else return sqrt(n) + f(n - 1);
}
double fun(int n)
{
    double S=0;
    for (int i = 1; i<=n; i++)
    {
        S += f(i);
    }
    return S;
}

 //也可以使用pow函数。功能：求x的y次方。用法：pow(x,y)。