网易云下载进u盘:1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2003)(b+2003)
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1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2003)(b+2003)
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.