小学六年级的题,请高手用小学生看得懂的方法计算,谢谢了
...1\/(1+2)+1\/(1+2+3)+1\/(1+2+3+4)+1\/(1+2+3+4+5)...1\/(1+2+3+4+...
所以 1\/(1+2)=2\/(2*3)=2\/2-2\/3 1\/(1+2+3)=2\/(3*4)=2\/3-2\/4 1\/(1+2+3+4)=2\/(4*5)=2\/4-2\/5 1\/(1+2+3+4+5)=2\/(5*6)=2\/5-2\/6 ...1\/(1+2+3+4+...99)=2\/(99*100)=2\/99-2\/100 所以 结果为2\/2-2\/100=49\/50 ...
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20...
1\/(1+2)+1\/(1+2+3)+1\/(1+2+3+4)+...+1\/(1+2+3+...+100)
1\/(1+2+3+……+n)=2\/n(n+1)所以 1\/(1+2)+1\/(1+2+3)+1\/(1+2+3+4)+...+1\/(1+2+3+...+100)=2\/2(2+1) + 2\/3(3+1) + …… + 2\/100(100+1)=2×[1\/2(2+1) + 1\/3(3+1) + …… + 1\/100(100+1)]=2×[1\/2-1\/3+1\/3-1\/4+……+1\/99...
1\/(1+2)+1\/(1+2+3)+1\/(1+2+3+4)+…+1\/(1+2+3+4+…+19)
1\/(1+2)+1\/(1+2+3)+1\/(1+2+3+4)+…+1\/(1+2+3+4+…+19)=[1+1\/(1+2)+1\/(1+2+3)+1\/(1+2+3+4)+1\/(1+2+3+...+5)]-1 =[2\/(1*2)+2\/(2*3)+2\/(3*4)+2\/(4*5)+2\/(5*6)]-1 =[(2\/1-1\/2)+(2\/2-2\/3)+(2\/3-2\/4)+(2\/4-2\/5...
1+1\/(1+2)+1\/1+2+3)+1\/(1+2+3+4)……1\/(1+2+3+4+……2009)怎么做
1+2+3+……+n=n(n+1)\/2,所以,1\/(1+2+3+……+n)=2\/n*(n+1)。原式=1+2\/2*3+2\/3*4+2\/4*5+……+2\/2009*2010 =1+2(1\/2*3+1\/3*4+1\/4*5+……+1\/2009*2010)=1+2*(1\/2-1\/3+1\/3-1\/4+1\/4-1\/5+……+1\/2009-1\/2010)=1+2*(1\/2-1\/2010)=1+...
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*...
数学计算。1+1\/(1+2)+1\/(1+2+3)+1\/(1+2+3+4)+……+1\/(1+2+3+……+1...
因为1=1*2\/2 1+2=2*3\/2 ...1+2+3+4+...2003=2003*2004\/2 所以 1+1\/(1+2)+1\/(1+2+3)+1\/(1+2+3+4)+...+1\/(1+2+3+4+...2003)=2(1\/1 - 1\/2)+2(1\/2 -1\/3) +2(1\/3-1\/4)+...+2(1\/2003-1\/2004)=2-2\/2004 =2-1\/1002 =2003\/1002 ...
分数1\/(1+2) + 1\/(1+2+3)+1\/(1+2+3+4)+...+
最后答:99\/101 每一项的分母可以看做等差数列的的前n项和,即 n*(n+1)\/2,那么每一项就等于 2\/n*(n+1),对每一项拆分 可以变成2*[(1\/n)-1\/(n+1)]那么原式=2*[1\/2-1\/3+1\/3-1\/4+1\/4...-1\/100+1\/100-1\/101]=1-2\/101 =99\/101 ...
计算巧算1+1\/(1+2)+1\/(1+2+3)+1\/(1+2+3+4)+……1\/(1+2+3+……+100...
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+...+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\/(...
1+1\/(1+2)+1\/(1+2+3)+1\/(1+2+3+4)...1\/(1+2+3...+2006)
1:1 2:1+2 3:1+2+3 4:1+2+3+4 很明显,当第n个式子时:n:1+2+3+4+···+n =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...