求Y=1-1/2+1/3-1/4+...-1/2*n 的前n项之和。
时间: 2023-11-20 18:05:51 浏览: 62
以下是两种求Y=1-1/2+1/3-1/4+...-1/2*n 的前n项之和的方法:
1. 使用C语言编写函数求解
```c
#include <stdio.h>
float baz(int k);
int main() {
int n;
printf("输入表达式中n的值:");
scanf("%d", &n);
printf("表达式的值=%f\n", baz(n));
}
float baz(int k) {
float sum = 0, t = 1.0, n = 1.0;
while (k > 0) {
k--;
sum = sum + t * 1.0 / n;
t = -1 * t;
n++;
}
return (sum);
}
```
2. 使用Python编写代码求解
```python
a = int(input("请输入n的值:"))
m = 0
for x in range(1, a + 1):
n = ((-1) ** (x - 1)) / x
m = m + n
print("前{}项和为{:.2f}".format(a, m))
```
相关问题
求Y=1-1/2+1/3-1/4+...-1/2*n 的前n项之
以下是两种C语言求Y=1-1/2+1/3-1/4+...-1/2*n的前n项之和的例子:
1.使用函数求解
```c
#include <stdio.h>
float baz(int k);
int main() {
int n;
printf("输入表达式中n的值:");
scanf("%d", &n);
printf("表达式的值=%f\n", baz(n));
}
float baz(int k) {
float sum = 0, t = 1.0, n = 1.0;
while (k > 0) {
k--;
sum = sum + t * 1.0 / n;
t = -1 * t;
n++;
}
return (sum);
}
```
2.使用循环求解
```c
#include <stdio.h>
int main() {
int i, n, flag = 1;
float s = 0;
printf("put in n:");
scanf("%d", &n);
for (i = 1; i <= n; i++) {
s += 1.0 / i * flag;
flag *= -1;
}
printf("%f\n", s);
return 0;
}
```
c语言求Y=1-1/2+1/3-1/4+...-1/2*n 的前n项之和
以下是两种C语言求Y=1-1/2+1/3-1/4+...-1/2*n 的前n项之和的例子:
1.使用for循环计算前n项和:
```c
#include <stdio.h>
int main() {
int n, flag = 1;
float s = 0;
printf("请输入n:");
scanf("%d", &n);
for (int i = 1; i <= n; i++) {
s += 1.0 / i * flag;
flag *= -1;
}
printf("前%d项和为%f\n", n, s);
return 0;
}
```
2.使用函数进行封装:
```c
#include <stdio.h>
float getsum(int n) {
int flag = 1;
float result = 0;
for (int i = 1; i <= n; i++) {
result += 1.0 / i * flag;
flag *= -1;
}
return result;
}
int main() {
int n = 0;
printf("请输入n:");
scanf("%d", &n);
printf("前%d项和为%f\n", n, getsum(n));
return 0;
}
```