(1-3/2*4)*(1-3/3*5)*(1-3/4*6)*(1-3/5*7)*...*(1-3/96*98)*(1-3/97*99)=?
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/03 12:48:12
![(1-3/2*4)*(1-3/3*5)*(1-3/4*6)*(1-3/5*7)*...*(1-3/96*98)*(1-3/97*99)=?](/uploads/image/z/3950655-15-5.jpg?t=%281-3%2F2%2A4%29%2A%281-3%2F3%2A5%29%2A%281-3%2F4%2A6%29%2A%281-3%2F5%2A7%29%2A...%2A%281-3%2F96%2A98%29%2A%281-3%2F97%2A99%29%3D%3F)
(1-3/2*4)*(1-3/3*5)*(1-3/4*6)*(1-3/5*7)*...*(1-3/96*98)*(1-3/97*99)=?
(1-3/2*4)*(1-3/3*5)*(1-3/4*6)*(1-3/5*7)*...*(1-3/96*98)*(1-3/97*99)=?
(1-3/2*4)*(1-3/3*5)*(1-3/4*6)*(1-3/5*7)*...*(1-3/96*98)*(1-3/97*99)=?
(1-3/2×4)×(1-3/3×5)×(1-3/4×6)×(1-3/5×7)×……×(1-3/96×98)×(1-3/97×99)=
首先明白:
1-[3/n*(n+2)]=[n*(n+2)-3]/[n*(n+2)]
=[n^2+2n-3]/[n*(n+2)]
=[(n-1)*(n+3)]/[n*(n+2)]
=[(n-1)/n]*[(n+3)/(n+2)]
这里的n为从2开始的自然数
所以,上式
=[(1/2)*(5/3)]*[(2/3)*(6/5)]*[(3/4)*(7/6)]*……*[(94/95)*(98/97)]*[(95/96)*(99/98)]*[(96/97)*(100/99)]
然后将每一个中括号中前面一项均提出来,再将每一个中括号中的后一项提出来
=[(1/2)*(2/3)*(3/4)*……*(94/95)*(95/96)*(96/97)]*[(5/3)*(6/5)*(7/6)*……*(98/97)*(99/98)*(100/99)]
再每个中括号内隔项约分
=(1/97)*(100/4)
=25/97