1\3+1\6+1\10+1\15+1\21+1\28+1\36+1\45+1\50+1\55

先提取公因数2,然后再进行裂项求和即可.原式=2*(1/6+1/12+1/20+1/30.+1/110+1/100)=2*[(1/2-1/3)+(1/3-1/4)+(1/4-1/5)+(1/5-1/6)+.+(1/10-1/11)+1/100]=2*[1/2-1/11+1/100] 消去中间项=2*461/1100...为什么？看不懂哎怎么得数又变了我再写详细点！稍等！因为1/6=1/2-1/3，1/12=1/3-1/4，1/20=1/4-1/5,1/42=1/6-1/7……原式=2*(1/6+1/12+1/20+1/30+1/42+1/56+1/72+1/90+1/100+1/110)=2*(1/6+1/12+1/20+1/30+1/42+1/56+1/72+1/90+1/110+1/100)=2(1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10+1/10-1/11+1/100)=2*(1/2-1/11+1/100)消去中间项=2*461/1100=461/550我发觉1/100无法消去，如果题目没有抄错就是上面的结果。之前看漏了1/50，所以算的是：1\3+1\6+1\10+1\15+1\21+1\28+1\36+1\45+1\55=2*(1/6+1/12+1/20+1/30+1/42+1/56+1/72+1/90+1/110)=2*(1/6+1/12+1/20+1/30+1/42+1/56+1/72+1/90+1/110)=2(1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10+1/10-1/11)=2*(1/2-1/11)消去中间项=2*9/22=9/11不明白可追问！请采纳！谢谢！