题目

1/1*2*3+1/2*3*4+1/3*4*5+...1/98*99*100如何计算

提问时间：2020-07-24

答案

1/(1*2*3)+1/(2*3*4)+1/(3*4*5)+……+1/(98*99*100)
=(1/2)*(4-3)/(3*4)+(1/3)*(5-4)/(4*5)+(1/4)*(6-5)/(5*6)+……+(1/98)*(100-99)*(99*100)
=(1/2)*(1/3-1/4)+(1/3)*(1/4-1/5)+(1/4)*(1/5-1/6)+……+(1/98)*(1/99-1/100)
=(1/2)*(1/3)-(1/2)*(1/4)+(1/3)*(1/4)-(1/3)*(1/5)+(1/4)*(1/5)-(1/4)*(1/6)+……+(1/98)*(1/99)-(1/98)*(1/100)
=[(1/2)*(1/3)+(1/3)*(1/4)+(1/4)*(1/5)+.+(1/98)*(1/99)]-[(1/2)(1/4)+(1/3)*(1/5)+(1/4)*(1/6)+.+(1/98)*(1/100)]
=[(3-2)/(2*3)+(4-3)/(3*4)+(5-4)/(4*5)+.+(99-98)/(98*99)]-(1/2)[(4-2)/(2*4)+(5-3)/(3*5)+(6-4)/(4*6)+.+(100-98)/(98*100)]
=[1/2-1/3+1/3-1/4+1/4-1/5+.+1/98-1/99]-(1/2)[1/2-1/4+1/3-1/5+1/4-1/6+1/5-1/7.+1/96-1/98+1/97-1/99+1/98-1/100]
=[1/2-1/99]-(1/2)[1/2-1/99-1/100]
=1/2-1/99-1/4+1/198+1/200
=(1/2-1/99+1/198)+1/200-1/4
=49/99-49/200
=(200*49-99*49)/(99*200
=101*49/(99*200)
=4949/19800
由于时间仓促,最后两步约分可以有误,望原谅.