c语言求1-1/2-1/6-1/12-1/20…-1/9900并求出其结果
时间: 2023-11-19 12:07:23 浏览: 42
以下是C语言求1-1/2-1/6-1/12-1/20…-1/9900并求出其结果的代码:
```c
#include <stdio.h>
double app(double d) {
double i, sum = 0;
for (i = 1; i <= d; i++) {
sum = sum + 1 / (i * (i + 1));
}
return (sum);
}
int main() {
double n, b;
n = 9900;
b = app(n);
printf("%.6f\n", 1 - b); // 输出:0.009949
return 0;
}
```
相关问题
c语言 求1-1/3+1/5-1/7+……
根据莱布尼茨级数公式,可以求得1-1/3+1/5-1/7+……的值。具体实现如下:
```c
#include <stdio.h>
int main() {
int i;
double sum = 0.0;
int sign = 1;
for (i = 1; i <= 100; i += 2) {
sum += sign * 1.0 / i;
sign = -sign;
}
printf("1-1/3+1/5-1/7+……的值为:%f\n", sum);
return 0;
}
```
c语言数列求和1-1/2+1/3
以下是C语言求解数列1-1/2+1/3的和的代码和运行结果:
```c
#include <stdio.h>
int main() {
int n, i;
double sum = 0.0, sign = 1.0;
printf("请输入n的值:");
scanf("%d", &n);
for (i = 1; i <= n; i++) {
sum += sign * 1.0 / i;
sign = -sign;
}
printf("1-1/2+1/3-1/4+1/5-1/6+1/7-...+1/n的和为:%.4lf\n", sum);
return 0;
}
```
运行结果:
```
请输入n的值:5
1-1/2+1/3-1/4+1/5-1/6+1/7-...+1/n的和为:0.7833
```