=1-1/2+1/2+1/3-1/3+1/4+1/4-1/5+……+1/99-1/100
=1-1/100
=99/100
数学问题提问1\/1*2+1\/2*3+1\/3*4+1\/4*5...+1\/98*99+1\/99*100=
1\/1*2+1\/2*3+1\/3*4+1\/4*5...+1\/98*99+1\/99*100 =1-1\/2+1\/2+1\/3-1\/3+1\/4+1\/4-1\/5+……+1\/99-1\/100 =1-1\/100 =99\/100
用简便方法计算1\/1*2+1\/2*3+```+1\/99*100
=1-1\/2+1\/2-1\/3+1\/3-1\/4+...+1\/99-1\/100 =1-1\/100 =99\/100 这种方法叫做裂项相消法。
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
一个数学问题
1\/1*2+1\/2*3+1\/3*4+…+1\/99*100= 解：1\/(1*2)=1\/1-1\/2 1\/(2*3)=1\/2-1\/3 1\/(3*4)=1\/3-1\/4 ::1\/(99*100)=1\/99-1\/100 所以原式=(1\/1-1\/2)+……+（1\/99-1\/100）=99\/100
求助数学题 1\/1*2+1\/2*3+1\/3*4+…+1\/99*100=?
这是一道典型的问题，用裂项法很简单 1\/1*2=1\/2=1-1\/2 1\/2*3=1\/6=1\/2-1\/3 1\/3*4=1\/12=1\/3-1\/4 ……1\/99*100=1\/99-1\/100 将上述等式相加：其中-1\/2与1\/2、-1\/3与1\/3……-1\/99与1\/99相互抵消，最后只剩下1-1\/100 所以，本题答案为99\/100 ...
数学题 1\/1*2+1\/2*3+1\/3*4+……1\/99*100+1\/100*101=? 速度好吗...
通过上述公式分解式子：1\/1-1\/2+1\/2-1\/3+1\/3-1\/4+...+1\/99-1\/100+1\/100-1\/101，除了第一项和最后项，其他每2项可以消去，最后化简为 1\/1-1\/101=1-1\/101
小学奥数题1\/(1*2)+1\/(2*3)+1\/(3*4)+...+1\/(99*100)
1\/(1*2)+1\/(2*3)+1\/(3*4)+1\/(4*5)+1\/(5*6)+……+1\/(98*99)+1\/(99*100)=1-1\/2+1\/2-1\/3+...+1\/99-1\/100 =1-1\/100 =99\/100
小学数学奥数六年级1\/1*2*3+1\/2*3*4+1\/3*4*5+...1\/98*99*100
解：1\/(1*2*3)+1\/(2*3*4)+1\/(3*4*5)+……+1\/(98*99*100)=(1\/2)*(4-3)\/(3*4)+(1\/3)*(5-4)\/(4*5)+(1\/4)*(6-5)\/(5*6)+……+(1\/98)*(100-99)*(99*100)=(1\/2)*(1\/3-1\/4)+(1\/3)*(1\/4-1\/5)+(1\/4)*(1\/5-1\/6)+……+(1\/98)*(1\/...
数学计算题:1\/1X2+1\/2X3+1\/3X4+...+1\/99x100=? 请写出计算过程~
1\/1*2=1\/1-1\/2 1\/2*3=1\/2-1\/3……所以原式=1-1\/2+1\/2-1\/3+1\/3-1\/4+……-1\/99+1\/99-1\/100=99\/100
...原题是这样的 1\/1*2*3+1\/2*3*4+...+1\/98*99*100
=(1\/n-1\/(n+1))*1\/(n+2)=1\/n*1\/(n+2)-1\/(n+1)*1\/(n+2)=1\/2*(1\/n-1\/(n+2))-1\/(n+1)+1\/(n+2)1\/1*2*3+1\/2*3*4+...+1\/98*99*100 =1\/2*(1\/1-1\/3)-1\/2+1\/3 +1\/2*(1\/2-1\/4)-1\/3+1\/4 +1\/2*(1\/3-1\/5)-1\/4+1\/5 +……+1\/...
