计算1/2+(1/3+2/3)+(1/4+2/4+3/4)+…+(1/2002+2/2002+…+2001/2002

1\/2+(1\/3+2\/3)+(1\/4+2\/4+3\/4)+……+(1\/50+2\/50+3\/50+……+49\/50) 十 ...
原式=1\/2+(1\/3+2\/3)+...+(1+2+...+n-1)\/n =1\/2+(1\/3+2\/3)+...+(n-1)n\/(2n)=1\/2+(1\/3+2\/3)+...+(n-1)\/2 =(1+2+...+49)\/2 =(1+49)*49\/4 =1225\/2;

1\/2+(1\/3+2\/3)+(1\/4+2\/4+3\/4)+…+(1\/2005+2\/2005+……2003\/2005+2004\/...
则原式=1\/2*(2-1)+1\/2*(3-1)+…+1\/2*(2005-1)=1\/2*(2-1+3-1+…+2005-1)=1\/2*(1+2+…+2004)=1\/2*(2004+1)*2004*1\/2 =1004505

1\/2+(1\/3+2\/3)+(1\/4+2\/4+3\/4)+……+(1\/50+2\/50+……49\/50)等于几?
原式 =1\/2+1+(1+1\/2)+（1+1）+（1+1\/2+1）+.+（1+1+1+.+1\/2+.+1+1）=1\/2*24+1*（1+1+2+2+3+3+4+4+5+5+6+6+.+24+24）=12+600 =612 自己心算的结果可能不对,思路就是这样

求1\/2+(1\/3+2\/3)+(1\/4+2\/3+3\/4)...+(1\/100+2\/100...99\/100)之值。
规律：第n组，分母为n+1，分子依次从1到n 解：考察一般组，第n组：1\/(n+1)+2\/(n+1)+...+n\/(n+1)=(1+2+...+n)\/(n+1)=[n(n+1)\/2]\/(n+1)=n\/2 本题共99组。1\/2+(1\/3+2\/3)+(1\/4+2\/3+3\/4)...+(1\/100+2\/100...99\/100)=1\/2+2\/2+3\/2+...+99\/...

1\/2+1\/(2+3)+1\/(2+3+4)...+1\/(2+3+4...+200) 简便计算
所以1\/(2+3+4+...+n)=2\/[(n-1)(n+2)]=2\/3[(n+2)-(n-1)]\/[(n+2)(n-1)]=2\/3[1\/(n-1)-1\/(n+2)]所以1\/2+1\/（2+3）+1\/（2+3+4)...+1\/(2+3+4...+200)=1\/2+2\/3(1\/2-1\/5+1\/3-1\/6+1\/4-1\/7+1\/5-1\/8+...+1\/196-1\/199+1\/197-1\/2...

1\/2+(1\/3+2\/3)+(1\/4+2\/4+3\/4)+…+(1\/2006+2\/2006+...+2004\/2006+2005\/...
1\/n+2\/n+3\/n+...(n-1)n=[1+2+3+...+(n-1)]\/n=(n-1)(1+n-1)\/2n=n(n-1)\/2n=(n-1)\/2 原式=1\/2+2\/2+3\/2+...+2005\/2=(1+2+3+4+...+2005)\/2=2005(1+2005)\/(2×2）=2005×2006\/4=1005507.5

1\/2+(1\/3+2\/3)+(1\/4+2\/4+3\/4)+...+(1\/40+2?40+...+38\/40+39\/40)用简 ...
这是一个数列求和问题，每个括号了的分子相加就行了，也就变成了1\/2+2\/2+...+39\/2=390

数学1\/2+(1\/3+2\/3)+(1\/4+2\/4+3\/4)+(1\/5+2\/5+3\/5+4\/5)+...+(1\/50+2...
1\/2=1\/2 （1\/3+2\/3）=2\/2 （1\/4+2\/4+3\/4）=3\/2 （1\/5+2\/5+3\/5+4\/5）=4\/2 ……（1\/50+2\/50+...+49\/50）=49\/2 所以原式为1\/2+2\/2+3\/2+4\/2……+49\/2=1225\/2

计算1\/2+(1\/3+2\/3)+(1\/4+2\/4+3\/4)+...+(1\/40+2\/40+...+39\/40)
1\/n+2\/n+……+(n-1)\/n =[1+2+……+(n-1)]\/n =[n(n-1)\/2]\/n =(n-1)\/2 所以 1\/2+(1\/3+2\/3)+(1\/4+2\/4+3\/4)+...+(1\/40+2\/40+...+39\/40)=(2-1)\/2+(3-1)\/2+……+(40-1)\/2 =(1+2+……+39)\/2 =[39*(39+1)\/2]\/2 =390 ...

1\/2+(1\/3+2\/3)+(1\/4+2\/4+3\/4)+...+(1\/100+2\/100...+98\/100+99\/100...
1\/3+2\/3=1\/2×（3-1）1\/4+2\/4+3\/4=1\/2×（4-1）...原式=1\/2×（2-1）+1\/2×（3-1）+1\/2×（4-1）+...+1\/2×（100-1）=1\/2×（1+2+3+...+99）=1\/2×（1+99）×99×1\/2 =1\/4×100×99 =25×99 =2475 ...