C语言求s = 1+1/3+...+(1*2*3*...*n)/(3*5*7*...*(2*n+1)) 函数问题
时间: 2023-11-11 13:03:16 浏览: 71
```c
#include<stdio.h>
double sum(int n)
{
double s=0,t=1;
int i;
for(i=1;i<=n;i++)
{
t=t*(2*i-1)/(2*i+1);
s=s+t;
}
return s+1;
}
int main()
{
int n;
printf("请输入n的值:");
scanf("%d",&n);
printf("s=%lf\n",sum(n));
return 0;
}
```
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c语言求s=1+1/1!+1/2!+1/3!+...+1/10!
以下是C语言求s=1+1/1!+1/2!+1/3!+...+1/10!的代码和注释解释:
```c
#include <stdio.h>
int main() {
int i, j;
double s = 0, factorial = 1; // 定义变量s和factorial,s用于存储求和结果,factorial用于存储阶乘
for (i = 1; i <= 10; i++) { // 循环10次,分别计算1/1!、1/2!、1/3!、...、1/10!
factorial *= i; // 计算i的阶乘
s += 1 / factorial; // 将1/i!加入到s中
}
printf("s = %lf\n", s); // 输出结果
return 0;
}
```
C语言求s = 1+1/3+...+(1*2*3*...*n)/(3*5*7*...*(2*n+1))
以下是C语言代码实现s = 1+1/3+...+(1*2*3*...*n)/(3*5*7*...*(2*n+1))的求解过程:
```c
#include <stdio.h>
int main() {
int n;
float sum = 1.0;
float temp = 1.0;
int i, j;
printf("请输入n的值:");
scanf("%d", &n);
for (i = 1; i <= n; i++) {
temp = 1.0;
for (j = 1; j <= i; j++) {
temp *= j;
}
sum += temp / ((2 * i + 1) * 1.0);
}
printf("s = %f", sum);
return 0;
}
```