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1/2*3+1/4*3+1/5*4……1/2018*1/2019

这个题目你先同分,在计算 1+1/2+1/3+1/4=(12+6+4+3)/12=25/12(1/2+1/3+1/4+1/5)=25/12+1/5=125/60+12/60=137/60(1+1/2+1/3+1/4+1/5)=1+137/60=197/60(1/3+1/4)=7/12原式=25/12*137/60 -197/60 *7/12 =1/12 *1/60 (25*137 -197*7) =1/12 *1/60*2046 =2046/720=1023/360

每一项可以变形为:1/1*2*3=1/2(1/1*2-1/2*3)1/2*3*4=1/2(1/2*3-1/3*4)1/3*4*5=1/2(1/3*4-1/4*5)所以1/1*2*3+1/2*3*4+1/3*4*5+……1/15*16*17=1/2(1/1*2-1/16*17)=135/544

1/2÷3+1/3÷4+1/4÷5+1/5÷6+1/6÷7 =(1/2-1/3)+(1/3-1/4)+(1/4-1/5)+(1/5-1/6)+(1/6-1/7) =1/2-1/7 =5/14 满意请采纳

1/1*2=1/1-2/11/2*3=1/2-1/31/3*4=1/3-1/4`````1/99*100=1/99-1/100故,1/1*2+1/2*3+1/3*4+1/4*5+1/5*6+……+1/98*99+1/99*100 =1/1-1/2+1/2-1/3+1/3+1/99-1/100 =1/1-1/100 =99/100懂?这叫“裂项相消法”

1/1*3+1/2*4+1/3*5+1/4*6+1/5*7+1/6*8+1/7*9+1/8*10+1/9*11 =(1/2)*[(1-1/3)+(1/2-1/4)+(1/3-1/5)+(1/4-1/6)+(1/5-1/7)+(1/6-1/8)+(1/7-1/9)+(1/8-1/10)+(1/9-1/11)] =(1/2)*[(1-1/11)+(1/2-1/10)] =(1/2)*(10/11+2/5) =5/11+1/5=36/55.

2*3/1+3*4/1+4*5/1.2018/1+2019/1=(1/2-1/3)+(1/3-1/4)+(1/4-1/5)++(1/2018-1/2019)=1/2-1/3+1/3-1/4+1/4-1/5++1/2018-1/2019=1/2-1/2019=2017/4038

(1+1/2)*(1+1/3)*(1+1/4)*…*(1+1/99)*(1+1/100)=3/2*4/3*5/4**100/99*101/100=101/2

1/1*2=1-1/2 1/2*3=1/2-1/3 1/3*4=1/3-1/4 …… …… 1/n*(n+1)=1/n-1/(n+1) 以上各式相加得 1/1*2+1/2*3+1/3*4+1/4*5+……+1/n*(n+1) =1-1/(n+1) 所以把200带入得 199/200

解析:1/(1*3)=(1/2)(1/1-1/3)1/(2*4)=(1/2)(1/2-1/4)1/(3*5)=(1/2)(1/3-1/5)..1/(2017*2019)=(1/2)*(1/2017-1/2019)上述各式相加,S=(1/2)[1+1/2-1/2018-1/2019)整理,得:详细见附图PS:说句实在的,我没有想到,它能解的出来.

1/1*2+1/2*3+1/3*4+1/4*5……1/199*200=(1-1/2)+(1/2-1/3)+(1/3-1/4)+(1/4-1/5)+ .+(1/199-1/200)=1-1/200=199/200

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