18、求1+1/(1*2)+1/(2*3)+...1/(n*(n+1))

求1+1/(1*2)+1/(2*3)+...1/(n*(n+1))

程序代码:

/*
	时间:2017年6月30日16:49:25
	功能:求1+1/(1*2)+1/(2*3)+...1/(n*(n+1))
*/
# include <stdio.h>

int main(void)
{
	 int n, i;
	 double s = 0, sum = 0;
 
	 printf("请输入n的值:");
	 scanf("%d", &n);
	 for (i=1; i<=n; ++i)	//i值初始值为0,当n为0时,循环不执行;
	 {
		  s = 1.0 / (i*(i+1));
		  sum = sum + s;	//sum是用来存储1/(1*2)+1/(2*3)+...1/(n*(n+1)的值
	 }
	 sum += 1;				// 以循环内部sum的最终值+1,得出题中的解;

	 printf("%lf
", sum);
 
 return 0;
}

  

原文地址:https://www.cnblogs.com/wxt19941024/p/6546728.html