1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2003)(b+2003)
问题描述：
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2003)(b+2003)
答
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)]
把能抵消的抵消．