(1－1/2+1/3－1/4+1/5－1/6+1/7－1/8)/(1/5+1/6+1/7+1/8)

1-1\/2+1\/3-1\/4+1\/5-1\/6+1\/7-1\/8 前n项和
解答：Sn=1-1\/2+1\/3-1\/4+1\/5-1\/6+1\/7-1\/8+.+1\/(2n-1)-1\/2n 没有求和公式，但是如果 n 趋于 +∞ 时，lim（n->∞) sn = ln2 如果一个数列{an}，与首末项等距的两项之和等于首末两项之和，可采用把正着写与倒着写的两个和式相加，就得到一个常数列的和。

1-1\/2+1\/3-1\/4+1\/5-1\/6+1\/7-1\/8+1\/9-1\/10+1\/11-1\/12
一就是淡定慢慢算，，可以求数列1\/n的和与1\/2n的和作差吧

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...
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~...

简算(1-1\/2+1\/3-1\/4+1\/5-1\/6+...+1\/49-1\/50)\/(1\/26+1\/27+...+1\/50...
=[1\/4+1\/2*（1\/3+1\/5+...+1\/25）+（1\/27+1\/29+……1\/49）]\/（1\/26+1\/27+...+1\/50）

(1-1\/2+1\/3-1\/4+...+1\/2011-1\/2012)\/(1\/(1+2013)+1\/(2+2014)+...1\/...
用B1来表示1\/2+1\/4+1\/6+...1\/2010+1\/2012,,B2来表示(1+1\/2+1\/3+...1\/1005+1\/1006)则B=B1-B2 而A+B1=1+1\/3+1\/5+1\/7+...+1\/2009+1\/2011 + 1\/2+1\/4+1\/6+...+1\/2010+1\/2012 =1+1\/2+1\/3+...1\/1005+1\/1006+1\/1007+1\/1008+...+1\/2010+1\/201...

1\/2-1\/3+1\/4-1\/5+1\/6-1\/7+1\/8。。。计算有什么规律没有,谢谢?
例如，1\/2-1\/3+1\/4-1\/5=(4*5-2*3)\/2*3*4*5;1\/6-1\/7+1\/8-1\/9=(8*9-6*7)\/6*7*8*9;则1\/2-1\/3+1\/4-1\/5+1\/6-1\/7+1\/8-1\/9=[(4*5-2*3)*6*7*8*9-(8*9-6*7)*2*3*4*5]\/9!=(48*1457)\/9!=1457\/7560.如果必须计算出结果，目前而言只能是根据1...

数列计算 1\/2+1\/3+1\/4+1\/5+1\/6+1\/7+1\/8+1\/9=?
这个级数是发散的。简单的说，结果为∞ －－－ 用高中知识也是可以证明的，如下：1\/2≥1\/2 1\/3＋1\/4＞1\/2 1\/5＋1\/6＋1\/7＋1\/8＞1\/2 ……1\/[2^(k-1)+1]＋1\/[2^(k-1)+2]＋…＋1\/2^k＞[2^(k-1)](1\/2^k)＝1\/2 对于任意一个正数a，把a分成有限个1\/2 必然能够...

1\/2+1\/3+1\/4+1\/5+1\/6+1\/7+1\/8=?
1\/2+1\/3+1\/4+1\/5+1\/6+1\/7+1\/8=?=（1\/2+1\/4+1\/8）+（1\/3+1\/6）+（1\/5+1\/7）=7\/8+1\/2+12\/35 =11\/8+12\/35 =1又201\/280

1+1\/2+1\/3+1\/4+1\/5+1\/6+1\/7+1\/8+...1\/n等于多少?
ln(1+1\/x) = 1\/x - 1\/2x^2 + 1\/3x^3 - ...于是：1\/x = ln((x+1)\/x) + 1\/2x^2 - 1\/3x^3 + ...代入x=1,2,...,n，就给出：1\/1 = ln(2) + 1\/2 - 1\/3 + 1\/4 -1\/5 + ...1\/2 = ln(3\/2) + 1\/2*4 - 1\/3*8 + 1\/4*16 - ...1\/n = ...

C语言编程 逻辑运算 1+1\/2-1\/3+1\/4-1\/5+1\/6-1\/7+1\/8-1\/9+1\/10 在线...
for(i=2;i<=n;i++) \/\/从第二项开始，符号变换 { item=mark*1.0\/i; \/\/每一项 sum+=item; \/\/加到一起 mark=-mark; \/\/每一项的正负号变换 } return sum;} int main(){ printf("1+1\/2-1\/3+1\/4-1\/5+1\/6-1\/7+1\/8-1\/9+1\/10=%lf\\n",func(10));} ...