参考资料:祝你新年快乐!进步!
1\/1*2+1\/2*3+1\/3*4+...+1\/2002*2003+1\/2003*2004
=1-1\/2+1\/2-1\/3+1\/3+1\/4+...+1\/2003-1\/2004 =1-1\/2004 =2003\/2004 参考资料:祝你新年快乐!进步!
...求1\/ab+1\/(a+1)(b+1)+1\/(a+2)(b+2)+.+1\/(a+2002)(b+2002)
互为相反数 所以ab-2=0,b-1=0,得出a=2,b=1 原式=1\/1*2+1\/2*3+1\/3*4+.+1\/2002*2003+1\/2003*2004 =[(1\/1-1\/2)+(1\/2-1\/3)+(1\/3-1\/4)+...+(1\/2002-1\/2003)+(1\/2003+1\/2004)]=1\/1+1\/2004 =1+1\/2004 =2005\/2004 哎呀,搞得我头都晕了 ...
1\/1×2+1\/2×3+1\/3×4+……+1\/2002×2003.
= (1\/1 - 1\/2) +(1\/2 - 1\/3) + (1\/3 - 1\/4) +... + (1\/2002 -1\/2003 )= 1 - 1\/2003 = 2002\/2003
1\/1x2+1\/2x3+1\/3x4...1\/2002x2003 如题,
1\/1x2+1\/2x3+1\/3x4...1\/2002x2003 =(1-1\/2)+(1\/2-1\/3)+(1\/3-1\/4)+.+(1\/2002-1\/2003)=1-1\/2003 =2002\/2003
1\/(1*2)+1\/(2*3)+1\/(3*4)+...+1\/(2002*2003)=\/
这是数列问题(这里这是通常所说数列的一部分),首先找通项an=1\/n(n+1)=1\/n+1\/(n+1)总和S=1-1\/2+1\/2-1\/3+...+1\/(n-1)-1\/n+1\/n-1\/(n+1)=1-1\/(n+1)在这里,n=2002,把它代入上式计算就是了.答案是S=2002\/2003 ...
从1\/1*2+1\/2*3+...+1\/2002*2003怎么计算
先帮你分析一下:1\/1*2=1-1\/2,1\/(2*3)=1\/2-1\/3……1\/2002*2003=1\/2002-1\/2003记住这个式子:1\/(m*n)=1\/m-1\/n,m、n为正整数,所以原式=1\/1-1\/2+1\/2-1\/3+1\/3-1\/4+…-1\/2002+1\/2002-1\/2003=1-1\/2003=2002\/2003,明白了没?
...解答1\/1x2+1\/2x3+...+1\/2002x2003+1\/2003x2004+1\/2004x2005
原式 =1-1\/2+1\/2-1\/3+...1\/2002-1\/2003+1\/2003-1\/2004+1\/2004-1\/2005 =1-1\/2005 =2004\/2005
一乘二分之一加二乘三分之一加三乘四分之一加...2003乘2004分之一的...
2、因为:1\/(1×2)=1-1\/2;1\/(2×3)=1\/2-1\/3;1\/(3×4)=1\/3-1\/4;1\/(4×5)=1\/4-1\/5;1\/(5×6)=1\/5-1\/6。………1\/(2002×2003)=1\/2002-1\/2003 1\/(2003×2004)=1\/2003-1\/2004 一乘二分之一加二乘三分之一加三乘四分之一加...2003乘2004分之一...
1\/1*2+1\/2*3+1\/3*4+1\/4*5...+1\/99*100怎么用简便方法计算
1\/1*2+1\/2*3+1\/3*4+1\/4*5...+1\/99*100 =1-1\/2+1\/2-1\/3+1\/3-1\/4+...+1\/99-1\/100 =1-1\/100 =99\/100
1*2\/1+2*3\/1+3*4\/1……2002*2003\/1答案和过程
1\/1×2+1\/2×3+1\/3×4+……+1\/2002×2003 =1-1\/2+1\/2-1\/3+1\/3-1\/4+……+1\/2002-1\/2003 =1-1\/2003 =2002\/2003.注:任意两个连续自然数的倒数的积,等于这两个连续自然数倒数的差.