(1+1/1*3)*(1+1/2*4)*(1+1/3*5)*(1+1/4*6)*…(1+1/98*100)*(1+1/99*100)

过程

(1+1/1*3)*(1+1/2*4)*(1+1/3*5)*(1+1/4*6)*…(1+1/98*100)*(1+1/99*100)

=[(2*2)/(1*3)]*[(3*3)/(2*4)]*[(5*5)/(4*6)]*...*[(99*99)/(98*100)]*[9901/(99*100)]

注意到"中间的每一项",其分子可与前后项约分完(第一个分母与前一项的分子约分,第二个分母与后一个分子约分),所以上式="第一项剩余2"*"倒数第二项剩余(1/100)"*"最后一项9901/(99*100)

=2*(1/100)*(9901/99/100),

=50*(9901/99/100)

=9901/198,估计所给题目的最后两项,被您弄错了,以致结果复杂,如果改成如下的则要好办得多!

(1+1/1*3)*(1+1/2*4)*(1+1/3*5)*(1+1/4*6)*…(1+1/97*99)*(1+1/98*100)

=[(2*2)/(1*3)]*[(3*3)/(2*4)]*[(5*5)/(4*6)]*...*[(98*98)/(97*99)]*[99/(98*100)]

=2*(99/100)

=99/50

=1又49/50

或=1.98