=1+(1-2/3)+(2/3-3/6)+(3/6-4/10)+...-2002/(1+2+3+...+2002)
=1+1-2/3+2/3-3/6+3/6-4/10+...-2002/(1+2+3+...+2002)
=1+1-2002/(1+2+3+...+2002)
=2-2002/(1+2+3+...+2002)
=2-2002/2005003
=2-2/2003
=1又2001/2003
1+1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+...+1/(1+2+3+...+2002)
=1+(1-2/3)+(2/3-3/6)+(3/6-4/10)+...-2002/(1+2+3+...+2002)最后一个分数前面为减号,得注意
要注意发现规律
1/(1+2)=+1-2/(1+2)
1/(1+2+3)=2/(1+2)-3(1+2+3)
1/(1+2+3+4)=3(1+2+3)-4(1+2+3+4)
...
=1+(1-2/3)+(2/3-3/6)+(3/6-4/10)+...-2002/(1+2+3+...+2002)
=1+1-2/3+2/3-3/6+3/6-4/10+...-2002/(1+2+3+...+2002)
=1+1-2002/(1+2+3+...+2002)
=2-2002/(1+2+3+...+2002)
=2-2002/2005003
=2-2/2003
=1又2001/2003
楼上的这位计算是正确的方法
=1+(1-2/3)+(2/3-3/6)+(3/6-4/10)+...-2002/(1+2+3+...+2002)
=1+1-2/3+2/3-3/6+3/6-4/10+...-2002/(1+2+3+...+2002)
=2-2002/(1+2+3+...+2002)
=2-2/2003
=1又2001/2003
1+1\/(1+2)+1\/(1+2+3)+...+1\/(1+2+3+...+2002)怎么用小学六年级学生的思...
=1+1-2002\/(1+2+3+...+2002)=2-2002\/(1+2+3+...+2002)=2-2002\/2005003 =2-2\/2003 =1又2001\/2003 1+1\/(1+2)+1\/(1+2+3)+1\/(1+2+3+4)+...+1\/(1+2+3+...+2002)=1+(1-2\/3)+(2\/3-3\/6)+(3\/6-4\/10)+...-2002\/(1+2+3+...+2002)最后一个分数前...
1+1\/(1+2)+1\/(1+2+3)...1\/(1+2+3+4+...+2004)
首先分析得知，分子都是1，分母则是一个等差数列的前n项和，一共是2004项求和，设A1=1,An=1\/(n*(1+n)\/2)=2\/(n*(n+1))=2\/n-2\/(n+1),1+1／（1+2）+1／（1+2+3）...1\/(1+2+3+4+...+2004)=(2\/1-2\/2)+(2\/2-2\/3)+(2\/3-2\/4)+...+(2\/2004-2\/2005)=2...
1+1\/(1+2)+1\/(1+2+3)+...+1\/1+2+3+...+2004=?
1+2=2*3\/2 1+2+3=3*4\/2 1+2+3+4=4*5\/2 1+2+3+……+2004=2004*2005\/2 所以，1+1\/(1+2)+1\/(1+2+3)+1\/(1+2+3+4)+...+1\/(1+2+3+...+2006)=1+2\/（2*3）+2\/（3*4）+2\/（4*5）+……+2\/（2004*2005）=2[（1\/2+1\/（2*3）+1\/（3*4）+1\/（...
1+1\/(1+2)+1\/(1+2+3)+…+1\/(1+2+3+…+n)这个怎么计算啊
因为1+2+3+..+n=n(n+1)\/2 所以[1\/（1+2+3+…+n)]=2\/n(n+1)=2[1\/n-1\/(n+1)]所以Sn=1+[1\/(1+2)]+〔1\/(1+2+3)〕+[1\/(1+2+3+4)]+……+[1\/(1+2+3+……+n)]=2[1\/1-1\/2]+2[1\/2-1\/3]+2[1\/3-1\/4]+...+2[1\/n-1\/(n+1)]=2[1-1\/2...
1+1\/(1+2)+1\/(1+2+3)+…+1\/(1+2+3+…100)=
第二种：因为：1+2=2*3\/2 1+2+3=3*4\/2 1+2+3+4=4*5\/2 1+2+3+……+100=100*101\/2 所以，1+1\/(1+2)+1\/(1+2+3)+1\/(1+2+3+4)+...+1\/(1+2+3+...+2006)=1+2\/（2*3）+2\/（3*4）+2\/（4*5）+……+2\/（100*101）=2[（1\/2+1\/（2*3）+1\/（3*...
1\/(1+2) +1\/(1+2+3) + 1\/(1+2+3+4) +...+1\/(1+2+3+...+2009) 怎么计算...
整个运算的每两个加号之间的为一个项，则总共有2008个项，其普通式为An，接下来我们先计算An的普通式，因为括弧里为分母，分母的普通式为n(n+1)\/2,求倒则为An=2\/n(n+1)，而2\/n(n+1)=2（1\/n - 1\/(n+1)），于是原式即为求普通式为An=2（1\/n - 1\/(n+1)）（n>1)，共有...
1+1\/(1+2)+1\/(1+2+3)+1\/(1+2+3+4)...1\/(1+2+3...+2006)
=n（n+1)\/2 1:1\/1 2:1\/（1+2)3:1\/（1+2+3)4:1\/ (1+2+3+4)很明显，当第n个式子时：n:1\/(1+2+3+4+···+n)=1\/[n（n+1)\/2]=2\/n(n+1)则原式=1+1\/(1+2)+1\/(1+2+3)+1\/(1+2+3+4)+···+1\/(1+2+3+···+2006)=1+2\/2×3+2\/3×4+2\/...
1+1\/(1+2)+1\/(1+2+3)+1\/(1+2+3+4)+…+1\/(1+2+3+…+100) 简便计算方法...
简便计算方法：1+2+3+...+n=n(n+1)\/21\/(1+2+3+...+n)=2\/n(n+1)=2[1\/n-1\/(n+1)]1+1\/(1+2)+1\/(1+2+3)+1\/(1+2+3+4)+...+1\/(1+2+3+...+100)=2[(1-1\/2)+(1\/2-1\/3)+(1\/3-1\/4)+...+(1\/100-1\/101)]=2(1-1\/101)=200\/101 它的原理是...
计算巧算1+1\/(1+2)+1\/(1+2+3)+1\/(1+2+3+4)+……1\/(1+2+3+……+100...
所以，1+1\/(1+2)+1\/(1+2+3)+1\/(1+2+3+4)+...+1\/(1+2+3+...+100)=1+2\/（2*3）+2\/（3*4）+2\/（4*5）+……+2\/（100*101）=2[（1\/2+1\/（2*3）+1\/（3*4）+1\/（4*5）+……+1\/（100*101）〕因为：1\/（2*3）=1\/2-1\/3；1\/（3*4）=1\/3-1\/4；...
...1+1\/(1+2)+1\/(1+2+3)+1\/(1+2+3+4)+...+1\/(1+2+3+L+100)的计算过程...
如下：1+2+3+...+n=n(n+1)\/2 1\/(1+2+3+...+n)=2\/n(n+1)=2[1\/n-1\/(n+1)]1+1\/(1+2)+1\/(1+2+3)+1\/(1+2+3+4)+...+1\/(1+2+3+...+100)=2[(1-1\/2)+(1\/2-1\/3)+(1\/3-1\/4)+...+(1\/100-1\/101)]=2(1-1\/101)=200\/101 性质 若已知一个...
