求1/(1*2)+1/(2*3)+1/(3*4)+.+1/(N*(N+1))的值,N=20, 要求:按四舍五入的方式精确
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求1/(1*2)+1/(2*3)+1/(3*4)+.+1/(N*(N+1))的值,N=20, 要求:按四舍五入的方式精确到小数点后第二位.
1/(1*2)+1/(2*3)+1/(3*4)+.+1/(N*(N+1))
= 1-1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ...+ 1/N - 1/(N+1)
= 1- 1/(N+1) = N/(N+1)
所以程序是:
#include
#define N 20
int main()
{
float a = N;
float b = 0;
b = a/(a+1.0);
printf("%3.2f\n",b);
return 0;
}
= 1-1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ...+ 1/N - 1/(N+1)
= 1- 1/(N+1) = N/(N+1)
所以程序是:
#include
#define N 20
int main()
{
float a = N;
float b = 0;
b = a/(a+1.0);
printf("%3.2f\n",b);
return 0;
}
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