求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)..........1/(a+2007)(b+2007)的值

求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)..........1/(a+2007)(b+2007)的值

int a = ?, b = ?, sum = 0 ;

for(int i = 0 ; i <=2007 ; i ++ )

{

sum += 1.0/(a+i)/(b+i) ;

}

print sum

1/(a+i)(b+i)=1/(b-a)*[1/(a+i)-1/(b+i)] 所以∑1/(a+i)(b+i)=1/(b-a)*[∑1/(a+i)-∑1/(b+i)]

#include<stdio.h> int main() { int a,b,i; double res=0; scanf("%d%d",&a,&b); for(i=0;i<=2007;i++) res+=1.0/((a+i)*(b+i)); printf("%lf\n",res); return 0; }