v=1/2+1/3+1/4+1/5+1/6.
原式=(1-u)v-(1-v)u=v-u=1/6
设m=1/2+1/3+1/4+1/5
原式
=(1-m)*(m+1/6)-(1-m-1/6)*m
=m-m*m+1/6-1/6*m-m+m*m+1/6*m
=1/6
计算(1-1\/2-1\/3-1\/4-1\/5)(1\/2+1\/3+1\/4+1\/5+1\/6)-(1-1\/2-1\/3-1\/4-1...
设u=1\/2+1\/3+1\/4+1\/5，v=1\/2+1\/3+1\/4+1\/5+1\/6.原式=(1-u)v-(1-v)u=v-u=1\/6
...1\/2+1\/3+1\/4+1\/5+1\/6)-(1-1\/2-1\/3-1\/4-1\/5-1\/6)乘以(1\/2+1\/3+...
画红线的它们相减等于0，六分之一是从后面的两个式子里提出来的
(1-1\/2-1\/3-1\/4-1\/5)(1\/2+1\/3+1\/4+1\/5+1\/6)-(1-1\/2-1
设（1-1\/2-1\/3-1\/4-1\/5）=X；（1\/2+1\/3+1\/4+1\/5）=Y;原式=X*(Y+1\/6)-(X-1\/6)Y=XY+X*1\/6-XY+Y*1\/6=1\/6*(X+Y)=1\/6*1=1\/6
计算:1-(1\/2-1\/3)-(1\/3-1\/4)-(1\/4-1\/5)-(1\/5-1\/6)-……-(1\/9-1\/10)
=1-1\/2+1\/3-1\/3+1\/4-1\/4+1\/5-1\/5……+1\/10 =1-1\/2+1\/10 =6\/10
(14)求算式直到1-1\/2+1\/3-1\/4+1\/5-1\/6+…第40项的和
1-1\/2+1\/3-1\/4+1\/5-1\/6+…第40项 =(1-1\/2)+(1\/3-1\/4)+(1\/5-1\/6)+...+(1\/39-1\/40)=1\/(1×2)+1\/(3×4)+1\/(5×6)+...1\/(39×40)=1-1\/(39+1)=1-1\/40 =39\/40
(1-1\/2+1\/3-1\/4+1\/5-1\/6+...+1\/1997-1\/1998)\/(1\/2000+1\/2002+1\/2004+...
分子可作如下变换：（1-1\/2+1\/3-1\/4+……+1\/1997-1\/1998+1\/1998）=(1+1\/2+1\/3+1\/4+…+1\/1998)-(1\/2+1\/4+1\/6+……+1\/1996+1\/1998)×2 =(1+1\/2+1\/3+1\/4+…+1\/1998)-(1+1\/2+1\/3+……+1\/998+1\/999)=1\/1000+1\/1001+1\/1002+……+1\/1998 分母可作如下...
...1 - 1\/2 + 1\/3 - 1\/4 + 1\/5 - 1\/6 + ... 现在请你求出该多项式的前n...
include<stdio.h> main（）into，n，l，t；for（n=1；n<=l；n++）t*=n；\/\/以各项分母相乘的结果来做公分母。o=t；\/\/fenzi printf（"1"，n）；for（n=2；n<=l；n++）if（n%2==1）o+=t\/n；printf（"+1\/%d"，n）；printf（"==%d\/%d==%f"，o，t，（double）o\/t）简介 ...
怎样求:1-1\/2+1\/3-1\/4+1\/5-1\/6+……+1\/(2n-1)-1\/(2n)的和?
当n=>∞时 S=ln2 1-1\/2+1\/3-1\/4……+1\/2n =1+1\/2+1\/3+1\/4……+1\/2n-2(1\/2+1\/4+……+1\/2n)=1\/(n+1)+1\/(n+2)+……1\/2n =1\/n(1\/(1+1\/n)+1\/(1+2\/n)+……+1\/(1+n\/n)=1\/(1+x)[从0积到1]=ln2 ...
C语言编写函数求:1-1\/2+1\/3-1\/4+1\/5-1\/6+...1\/n的值怎么做?结果是什么...
cpp文件，例如：test.cpp。2、在test.cpp文件中，输入C语言代码：。int n = 100;double s;for (int i = 1; i <= n; i++)if (i % 2 == 0)s -= 1.0 \/ i;else s += 1.0 \/ i;printf("%lf", s);3、编译器运行test.cpp文件，此时成功输出了摆动符号多项式的结果。
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\/1...
相减的项也写在一块，就能看出来了。原式 = 1 + 1\/3 + 1\/4 + 1\/5 + 1\/6 + 1\/7 + 1\/8 + 1\/9 + 1\/10 - 1\/2 - 1\/3 - 1\/4 - 1\/5 - 1\/6 - 1\/7 - 1\/8 - 1\/9 = 1 + 1\/10 - 1\/2 = 1\/2 + 1\/10 = 3\/5 希望对你有帮助，新年快乐~...
