如题所述
#include <stdio.h>
void main()
{
int i,n;
double s=0,sum=0;
printf("input n:");
scanf("%d",&n);
for (i=1;i<=n;i++)
{
sum+=i;
s+=1.0/sum;
}
printf("%lf\n",s);
}
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第1个回答 2014-07-05
1+ 1/((1+2)*2/2)+1/((1+3)*3/2)+......+1/((1+n)*n/2)=
2/(1*2)+2/(2*3)+2/(3*4)+......+2/(n*(n+1))=
2*(1-1/2+1/2-1/3+1/3-1/4......+1/n-1/(n+1))=
2*(1-1/(n+1))
原式等价于2*n/(n+1)
直接算吧
2/(1*2)+2/(2*3)+2/(3*4)+......+2/(n*(n+1))=
2*(1-1/2+1/2-1/3+1/3-1/4......+1/n-1/(n+1))=
2*(1-1/(n+1))
原式等价于2*n/(n+1)
直接算吧