=1/2(1+1/2+1/3+1/4)-1/3(1+1/2+1/3+1/4)+1/4(1+1/2+1/3+1/4)-1/5(1+1/2+1/3+1/4)
=(1+1/2+1/3+1/4)(1/2-1/3+1/4-1/5)
=(12+6+4+3)/12*(30-20+15-12)/60
=25/12*13/60
=65/144本回答被提问者和网友采纳
简便计算 (1\/2+1\/4+1\/6+1\/8)-(1\/3+1\/6+1\/9+1\/12)+(1\/4+1\/8+1\/12+1...
(1\/2+1\/4+1\/6+1\/8)-(1\/3+1\/6+1\/9+1\/12)+(1\/4+1\/8+1\/12+1\/16)-(1\/5+1\/10+1\/15+1\/20）=1\/2（1+1\/2+1\/3+1\/4）-1\/3（1+1\/2+1\/3+1\/4）+1\/4（1+1\/2+1\/3+1\/4）-1\/5（1+1\/2+1\/3+1\/4）=（1\/2-1\/3+1\/4-1\/5）（1+1\/2+1\/3+1\/4）=...
(1\/2+1\/4+1\/6+1\/8)-(1\/3+1\/6+1\/9+1\/12)+(1\/4+1\/8+1\/12+1\/16)-(1\/5...
原式=(1\/2)(1+1\/2+1\/3+1\/4)-(1\/3)(1+1\/2+1\/3+1\/4)+(1\/4)(1+1\/2+1\/3+1\/4)-(1\/5)(1+1\/2+1\/3+1\/4)=(1\/2-1\/3+1\/4-1\/5)(1+1\/2+1\/3+1\/4)=(13\/60)*(25\/12)=65\/144
...+1\/3+1\/4+1\/5+1\/6+1\/7+1\/8+1\/9+1\/10+1\/11+1\/12+1\/13+1\/14+1\/15+...
这是调和数列，没有简便方法 定义1：自然数的倒数组成的数列,称为调和数列. 定义2：若数列{an}满足1\/a（n+1）-1\/an=d（n∈N*，d为常数），则称数列{an}调和数列 人们已经研究它几百年了.但是迄今为止没有能得到它的求和公式只是得到它的近似公式(当n很大时): 1+1\/2+1\/3+...+1\/n≈...
c语言求1+1\/2+1\/4+1\/6+1\/8+1\/10+1\/12+1\/14+1\/16+1\/18+1\/20?
int i;float t,s=1;for (i=1;i<=10;i++){ t = 1.0\/(2*i);s += t;printf("%d %.4f %.4f\\n",i,t,s);} printf("%f\\n",s);
1+1\/2+1\/3+1\/4+1\/5+1\/6+1\/7+1\/8+1\/9+1\/10=多少
计算1+1\/2+1\/3+1\/4+1\/5+1\/6+1\/7+1\/8+1\/9+1\/10的过程如下：首先，我们可以通过拆分求和的方法简化计算：= 1 + (1\/2 + 1\/3 + 1\/6) + (1\/4 + 1\/5 + 1\/10) + 1\/7 + 1\/8 + 1\/9 = 1 + 1 + (1\/2 + 1\/3 - 1\/6) + (1\/4 + 1\/5 + 1\/10)= 1 +...
小学数学1\/2+1\/3+1\/4+1\/5+1\/6+1\/7+1\/8+1\/9+1\/10=? 1\/2+1\/3+1\/4+1\/...
1\/2+1\/3+1\/4+1\/5+1\/6+1\/10+1\/9+1\/8+1\/7 = 30\/60+ 20\/60+ 15\/60+ 12\/60+ 10\/60+ 6\/60+ 8\/72+ 9\/72+ 1\/7 = 50\/60+ 27\/60+ 16\/60 +17\/72 +1\/7 = 93\/60+ 85\/360+ 1\/7 = 558\/360+ 85\/360+ 1\/7 = 643\/360+ 360\/2520 = 4501\/2520+ 360\/2520 =...
...+1\/3+1\/4+1\/5+1\/6+1\/7+1\/8+1\/9+1\/10+1\/11+1\/12+1\/13+1\/14+1\/15...
首先提取公因式，归纳一下，就可以化成1\/15*（1+2+...+14）+1\/14*（1+2+...+13）+...1\/3*（1+2）+1\/2 然后根据括号内可以用等差求和的方法每个单项式归纳为1\/n*n\/2*(n-1)=(n-1)\/2 所以将N分别取15到2 提取公因式1\/2 最后变成（1+3+4+...+14）\/2=52.5 ...
1\/2+1\/3+1\/4+1\/5+1\/6+1\/7+1\/8+1\/9+1\/10+...1\/100这道题怎样简算_百度知 ...
没有，这是调和数列，很早就有数学家研究，比如中世纪后期的数学家Oresme在1360年就证明了这个级数是发散的。他的方法很简单： 1 +1\/2+1\/3 +1\/4 + 1\/5+ 1\/6+1\/7+1\/8 +... 1\/2+1\/2+(1\/4+1\/4)+(1\/8+1\/8+1\/8+1\/8)+... 注意后一个级数每一项对应的分数都小于调和级数...
1+1\/2+1\/3+1\/4+1\/5+1\/6+1\/7+1\/8+1\/9+1\/10。计算结果。
1+1\/2+1\/3+1\/4+1\/5+1\/6+1\/7+1\/8+1\/9+1\/10 =1+(1\/2+1\/3+1\/6)+(1\/4+1\/5+1\/10)+1\/7+1\/8+1\/9 =1+1+11\/20+1\/8+1\/7+1\/9 =1+1+27\/40+1\/9+1\/7 =1+1+283\/360+1\/7 =1+1+2341\/2520 =1+4861\/2520 ≈2.92896 ...
1\/2+1\/3+1\/4+1\/5+1\/6+1\/7+1\/8+1\/9+1\/10+1\/11.+1\/32答案是多少? 速求...
此题看似简单,实则牵涉到欧拉常数等高等数学的内容,非常复杂,难以计算.不过,因为所给32个数据均不超过1\/2,可以肯定的是,答案一定是一个小于16的数字.
