1/(2*4)=1/2*(1/2-1/4)
后面的都能这么化简
所以原式等于1/2*(1-1/3+1/3-1/5+1/5-1/7......+1/17-1/19)+1/2*(1/2-1/4+1/4-1/6......+1/18-1/20)
中间的能全部抵消 所以等于1/2*(1-1/19)+1/2*(1/2-1/20)
=9/19+9/40
=531/760
式中消去相抵消的得：
1／2（1+1／2-1／19-1／20)=.....自己算吧..
1\/(1×3)+1\/(2×4)+1\/(3×5)+1\/(4×6)...1\/(18×20)
1\/(1*3)=1\/2*(1-1\/3)1\/(2*4)=1\/2*(1\/2-1\/4)后面的都能这么化简 所以原式等于1\/2*(1-1\/3+1\/3-1\/5+1\/5-1\/7...+1\/17-1\/19)+1\/2*(1\/2-1\/4+1\/4-1\/6...+1\/18-1\/20)中间的能全部抵消 所以等于1\/2*(1-1\/19)+1\/2*(1\/2-1\/20)=9\/19+9\/40 =531\/...
...4)+1\/(3×5)+...+1\/[n(n+2)]=3\/4-(2n+3)\/[2(n+1)(n+2)]
呵呵，这是数学的一种方法，叫做“裂项相消法”，常用于分式的相加或相乘中，可以写 1\/（1*3）+1\/（2*4）+1\/（3*5）+．．．+1\/n(n+2)=[(1-1\/3)+(1\/2-1\/4)+(1\/3-1\/5)...+(1\/<n-1>-1\/<n+1>)+(1\/n+1\/<n+2>)]\/2 =(1+1\/2-1\/<n+1>-1\/<n+2>)\/2 ...
1*1\/3+2*1\/4+3*1\/5+4*1\/6+...+19*1\/21怎么做啊,感谢
=19-2（1\/3+1\/4+1\/5+1\/6+……+1\/21），1\/3+1\/4+1\/6+1\/12=（4+3+2+1）\/12=5\/6，1\/7+1\/14+1\/21=（6+3+2）\/42=11\/42，1\/5+1\/8+1\/10+1\/16+1\/20=（16+10+8+5+4）\/80=43\/80，1\/9+1\/15+1\/18=（10+6+5）\/90=7\/30 5\/6+11\/42+43\/80+7\/30+1\/...
...1\/(1*2*3)+1\/(2*3*4)+1\/(3*4*5)+...1\/(18*19*20)=?
1\/(1*2*3)=(1\/2)*(1\/1*2-1\/2*3)1\/(2*3*4)=(1\/2)*(1\/2*3-1\/3*4)1\/(3*4*5)=(1\/2)*(1\/3*4-1\/4*5)以此类推，规律就出现了。接下来就看你了。
数学题:1\/(1*3)+1\/(2*4)+1\/(3*5)+...+1\/[n*(n+2)]=
1\/(1*3)=(1-1\/3)\/2 1\/(2*4)=(1\/2-1\/4)\/2 ...原式*2= 1-1\/3+1\/2-1\/4...1\/n-1\/(n+2)=1+1\/2-1\/3+1\/3-1\/4+1\/4...1\/(n+1)-1\/(n+2)=1+1\/2-1\/(n+1)-1\/(n+2)=3\/2-1\/(n+1)-1\/(n+2)然后把让面的结果除以二化简一下就行了。
计算:1\/(1*3)+1\/(2*4)+1\/(3*5)+…+1\/(n(n+1)(n+2))
则：1\/(1×3)=(1\/1-1\/3)×1\/2 1\/(2×4)=(1\/2-1\/4)×1\/2 1\/(3×5)=(1\/3-1\/5)×1\/2 ……1\/[n(n+2)]=[1\/n-1\/(n+2)]×1\/2 那么：1\/(1×3)+1\/(2×4)+1\/(3×5)+……+1\/[n(n+2)]=(1\/1-1\/3)×1\/2+(1\/2-1\/4)×1\/2+(1\/3-1\/5)×1\/...
一道初中数学计算题:1\/1*3+1\/2*4+1\/3*5+1\/4*6+1\/5*7+1\/6*8---1\/17...
=1\/2[(1-1\/3)+(1\/2-1\/4)+(1\/3-1\/5)+...+(1\/17-1\/19)+(1\/18-1\/20)=1\/2[1+1\/2-1\/3+1\/3-1\/4+1\/4-...-1\/18+1\/18-1\/19-1\/20]=1\/2(1+1\/2-1\/19-1\/20)=(1\/2)*(531\/380)=531\/760
已知A=1\/(1×2)+1\/(3×4)+1\/(5×6)+……+1\/(1999×2000),B=1\/1000+...
A＞B，理由如下所示：
1\/1×3+1\/2×4+1\/3×5+…+1\/9×11+1\/10×12。 有点乱,\/是分数分号
\/2 = (1\/1 - 1\/3 + 1\/3 - 1\/5 + 1\/5 - 1\/7 + 1\/7 - 1\/9 + 1\/9 - 1\/11)\/2 + (1\/2 - 1\/4 + 1\/4 - 1\/6 + 1\/6 - 1\/8 + 1\/8 - 1\/10 + 1\/10 - 1\/12)\/2 = (1 - 1\/11)\/2 + (1\/2 - 1\/12)\/2 = 5\/11 + 5\/24 = 175\/264 ...
1\/(1+3)+1\/(3+5)+1\/(5+7)+...1\/(2003+2005)
解：观察上式可知，所求的是从1至2005之间连续两个奇数和的倒数之和 ∵连续两个奇数可以表示为（2n-1）、（2n+1）∴两者的和为2n-1+2n+1=4n ∵2n-1=2003，n=1002 ∴n为正整数，n∈（1，1002）依题意得，上式=∑1\/（4n）=1\/4+1\/4*2+1\/4*3+……1\/（4*1002）=1\/4*（1+1...
