1/2+1/6+1/12+1/20+1/30 简便计算~~

1\/2+1\/6+1\/12+1\/20+1\/30=简便计算。
1\/2+1\/6+1\/12+1\/20+1\/30简便计算结果为5\/6。解：1\/2+1\/6+1\/12+1\/20+1\/30 =1\/(1x2)+1\/(2x3)+1\/(3x4)+1\/(4x5)+1\/(5x6)=(1-1\/2)+(1\/2-1\/3)+(1\/3-1\/4)+(1\/4-1\/5)+(1\/5-1\/6)=1-1\/2+1\/2-1\/3+1\/3-1\/4+1\/4-1\/5+1\/5-1\/6 =1-1\/6 ...

1\/2+1\/6+1\/12+1\/20+1\/30=简便计算?
1\/2+1\/6+1\/12+1\/20+1\/30 =30\/60+10\/60+5\/60+3\/60+2\/60 =(30+10+5+3+2)\/60 =50\/60 =5\/6

“1\/2+1\/6+1\/12+1\/20+1\/30”怎样用简便方法计算?
简便计算 1\/2+1\/6+1\/12+1\/20+1\/30 =1\/（1x2）+1\/（2x3）+1\/（3x4）+1\/（4x5）+1\/（5x6）=1-1\/2+1\/2-1\/3+1\/3-1\/4+1\/4-1\/5+1\/5-1\/6 =1-1\/6 =5\/6

1\/2+1\/6+1\/12+1\/20+1\/30怎么简便算啊
解：1\/2+1\/6+1\/12+1\/20+1\/30 =1\/(1×2)+1\/(2×3)+1\/(3×4)+1\/(4×5)+1\/(5×6)=1\/1-1\/2+1\/2-1\/3+1\/3-1\/4+1\/4-1\/5+1\/5-1\/6 =1-1\/6 =1\/5

1\/2+1\/6+1\/12+1\/20+1\/30怎样用简便方法计算
根据基本裂项公式：可知：一、定义：裂项法，这是分解与组合思想在数列求和中的具体应用。是将数列中的每项（通项）分解，然后重新组合，使之能消去一些项，最终达到求和的目的。 通项分解（裂项）倍数的关系。二、其他裂项法公式：

...加12分之一加20分之一加30分之一 一步一步算, 简便计算
1\/2+1\/6+1\/12+1\/20+1\/30 =30\/60+10\/60+5\/60+3\/60+2\/60 =50\/60 =5\/6

1\/2+1\/6+1\/12+1\/20+1\/30+...?
1\/2+1\/6+1\/12+1\/20+1\/30+.+1\/9900简便运算过程如 1\/2+1\/6+1\/12+1\/20+1\/30+...+1\/9900 =1\/1*2+1\/2*3+1\/3*4+1\/4*5+1\/5*6+...+1\/99*100 =1-1\/2+1\/2-1\/3+1\/3-1\/4+1\/4-1\/5+1\/5-1\/6+1\/6-...-1\/99+1\/99-1\/100 =1-1\/100 =99\/100 所以1...

1\/2+1\/6+1\/12+1\/20+1\/30+1\/42的简便运算?
解题思路:常规思路，先通分，再求和。通分要找规律，分母方便通分的，放在一起进行计算。显然，1\/2，1\/6，1\/12这三个数分母通分都为12。1\/20，1\/30分母通分为60。所以，简便运算先算前三个数和最后一个数。原=1\/2+1\/6+1\/12+1\/42+1\/20+1\/30 =6\/12+2\/12+1\/12+1\/42+3\/60+2\/60...

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的简便运算。
1\/2+1\/6+1\/12+1\/20+1\/30+1\/42 =1\/(1x2)+1\/(2x3)+1\/(3x4)+1\/(4x5)+1\/(5x6)+1\/(6x7)=(1 -1\/2) +(1\/2-1\/3)+...+(1\/6-1\/7)=1 -1\/7 =6\/7 异分母分数相加：1、异分母分数相加，先通分，再按同分母分数相加法去计算，最后要化成最简分数。例1：3\/4+5\/7...