(1-1/2-1/3-1/4-1/5)(1/2+1/3+1/4+1/5+1/6)-(1-1/2-1/3-1/4-1/5-1/6)(1/2+1/3+1/4+1/5)的结果为
要详细解答过程，分析
则原式=(1-a)(a+1/6)-(1-a-1/6)*a
=a(1-a)+1/6*(1-a)-a(1-a)+1/6*a
=1/6*(1-a)+1/6*a
=1/6*1-1/6*a+1/6*a
=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
简算(1-1\/2+1\/3-1\/4+1\/5-1\/6+...+1\/49-1\/50)\/(1\/26+1\/27+...+1\/50...
（1-1\/2+1\/3-1\/4+1\/5-1\/6+...+1\/49-1\/50）\/（1\/26+1\/27+...+1\/50）=[（1-1\/2 ）+（1\/3+1\/5+...+1\/49）-（1\/4+1\/6+1\/8+……1\/50）]\/（1\/26+1\/27+...+1\/50）=[1\/2+（1\/3+1\/5+...+1\/49）-1\/2*（1\/2+1\/3+1\/4+……1\/25）]\/（1\/26...
(1-1\/2+1\/3-1\/4+1\/5-1\/6+...+1\/1997-1\/1998)\/(1\/2000+1\/2002+1\/2004+...
分母可作如下变换：(1\/2000+1\/2002+1\/2004+……+1\/3996)=1\/2*(1\/1000+1\/1001+1\/1002+……+1\/1998)所以 （1-1\/2+1\/3-1\/4+……+1\/1997-1\/1998+1\/1998)\/(1\/2000+1\/2002+1\/2004+……+1\/3996)=(1\/1000+1\/1001+1\/1002+……+1\/1998)\/[1\/2*(1\/1000+1\/1001+1\/1002+...
(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\/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。
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 ...
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
...1 - 1\/2 + 1\/3 - 1\/4 + 1\/5 - 1\/6 + ... 现在请你求出该多项式的前n...
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 在数学中，多项式（polynomial）是指由变量、系数...
