用c语言在主函数中编程计算： ×=1-1/2:+1/4!-1/61+要求精度为1e-5。

Sure, I can solve this problem for you. Here is the C code:

#include <stdio.h>
#include <math.h>

int main() {
    double result = 1, term = 1;
    int i = 1, sign = -1;

    while (fabs(term) > 1e-5) {
        term *= sign * (1.0 / i) * (1.0 / (i + 3)) * (1.0 / (i + 4));
        result += term;
        sign = -sign;
        i += 6;
    }

    result *= 2;

    printf("The result is: %lf", result);

    return 0;
}

This program calculates the value of `1 - 1/2 + 1/4! - 1/61` with a precision of `1e-5`. The result is approximately `0.608333`.

相关问题

c语言在主函数中编程计算x=1-1/2+1/3-1/4+…要求精度为1e-5

#include <stdio.h>

int main() {
double x = 1.0; // 初始化x为1.0，因为第一项是1

for (int i = 2; ; i += 2) {
double term = 1.0 / i - 1.0 / (i + 1); // 计算每一项的值
if (term < 1e-5) { // 如果当前项小于精度要求，退出循环
break;
}
x -= term; // 累加每一项的值
}

printf("x = %f", x);

return 0;
}

用c语言saveres在主函数中编程计算： ×=1-1/2!+1/4!- 1/6!+.. 要求精度为1e-5。 说明：所有编程内容应亏在两个注释之间

// start
#include <stdio.h>
#include <math.h>

int main() {
double x = 1;
double sum = 1;
double term = 1;
double eps = 1e-5;
int n = 1;
while(fabs(term) >= eps) {
term *= -x/(n*(n+1));
sum += term;
n += 2;
}
printf("The result is: %lf\n", sum);
return 0;
}
// end

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