1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72

如题所述

第1个回答  2019-05-17
1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72
=1/2+(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)
=1/2+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
=1-1/9
=8/9
第2个回答  2019-05-17
这个应该是一个查询,当开始日期不为空的时候开始查询between...and 语句就是查询的时候按这个范围内条件进行查询的SQL语句
第3个回答  2019-05-17
原式=1-1/2+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
=1-1/9
=8/9本回答被网友采纳
第4个回答  2012-06-11
1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72
=1-1/2+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
=1-1/9
=8/9

简便计算1\/2+1\/6+1\/12+1\/20+1\/30+1\/42+1\/56+1\/72
1\/2+1\/6+1\/12+1\/20+1\/30+1\/42+1\/56+1\/72 =1-1\/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 =1-1\/8 =7\/8 裂项相加

1\/2+1\/6+1\/12+1\/20+1\/30+1\/42+1\/56+1\/72+1\/90怎么简便计算?
1\/2+1\/6+1\/12+1\/20+1\/30+1\/42+1\/56+1\/72+1\/90=9\/10 方法:裂项相消法 1\/[n(n+1)]=(1\/n)- [1\/(n+1)]由题意得:1\/6=1\/[2(2+1)]、1\/12=1\/[3(3+1)]、1\/20=1\/[4(4+1)]、1\/30=1\/[5(5+1)]、依次可以表达为1\/[n(n+1)]的形式。所以可得:...

1\/2+1\/6+1\/12+1\/20+1\/30+1\/42+1\/56+1\/72= 简便计算,快!
1\/2+1\/6+1\/12+1\/20+1\/30+1\/42+1\/56+1\/72 =1\/(1*2)+1\/(2*3)+1\/(3*4)+...+1\/(8*9)=1-1\/2+1\/2-1\/3+...+1\/8-1\/9 =1-1\/9 =8\/9

数学:计算 1\/2+1\/6+1\/12+1\/20+1\/30+1\/42+1\/56+1\/72+1\/90=? (例:1\/2...
1\/2+1\/6+1\/12+1\/20+1\/30+1\/42+1\/56+1\/72+1\/90 =(1-1\/2)+(1\/2-1\/3)+(1\/3-1\/4)+...+(1\/8-1\/9)+(1\/9-1\/10)=1-1\/10 =9\/10

...的数学题:1\/2+1\/6+1\/12+1\/20+1\/30+1\/42+1\/56+1\/72求简便方法_百度知 ...
1\/2+1\/6+1\/12+1\/20+1\/30+1\/42+1\/56+1\/72 把1\/6项以后的各项都拆成两项,形成前后可以+-抵消的项,使计算简便 =1\/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\/2+1\/2-1\/9 =1-1\/9 =8\/9 ...

1\/2+1\/6+1\/12+1\/20+1\/30+1\/42+1\/56 +1\/72+1\/90=?【简便方法】_百度知 ...
原式=1\/(1*2)+1\/(2*3)+...+1\/(9*10)1\/(n*(n+1))=1\/n-1(n+1)所以 原式=1\/1-1\/2+1\/2-1\/3+1\/3-1\/4+...+1\/8-1\/9+1\/9-1\/10=1\/1-1\/10=9\/10

1\/2十1\/6十1\/12十1\/20十1\/30十1\/42十1\/56十1\/72.怎样做
1\/2+1\/6+1\/12+1\/20+1\/30+1\/42+1\/56+1\/72=8\/9 这样算是因为他有8个数,分子就是8,然后再用最后一个分数的分母除8就是分子

1\/2+1\/6+1\/12+1\/20+1\/30+1\/42+1\/56+1\/72怎么简便计算?
1\/2+1\/6+1\/12+1\/20+1\/30+1\/42+1\/56+1\/72 =1-1\/2+1\/2-1\/3+1\/3-1\/4+...+1\/7-1\/8+1\/8-1\/9 =1-1\/9 =8\/9 =9分之8 朋友,请【采纳答案】,您的采纳是我答题的动力,谢谢。

1\/2+1\/6+1\/12+1\/20+1\/30+1\/42+1\/56+1\/72=?
1\/2+1\/6+1\/12+1\/20+1\/30+1\/42+1\/56+1\/72 =1-1\/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+(-1\/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+0-1\/9 =1-1\/9 =8\/9 =9分...

...的数学题:1\/2+1\/6+1\/12+1\/20+1\/30+1\/42+1\/56+1\/72求简便方法_百度知 ...
1\/2+1\/6+1\/20+1\/30+1\/42+1\/56+1\/72 =1\/1 ×2+1\/2×3+1\/3×4……+1\/8×9 =(2-1)\/1 ×2+(3-2)\/2×3+(4-3)\/3×4……+(9-8)\/8×9 =1-1\/2+1\/2-1\/3+1\/3-1\/4+……+1\/8-1\/9 =1-1\/9 =8\/9 ....

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