C语言 实验10-5 递归求简单交错幂级数的部分和 (15分)

解法一，借用math库，直接根据题目即可写出

#include<math.h>
double fn (double x, int n)
{
	double sum;
	if (n == 1)
		sum = x;
	else
		sum = pow(x,n)*pow(-1,n-1) + fn(x,n-1);

	return sum;
}

解法二，比较难理解

double fn( double x, int n )
{
	double sum;
    if(n==1) 
    	sum = x;
    else 
		sum = x*( (-1) * fn(x,n-1) +1 );
    return sum;
}