1\/(1×2)+1\/(2×3)+1\/(3×4)...+1\/(48×49)+1\/(49×50)等于多少?
可以如下分析思考：1\/(1×2）+1\/(2×3）+1\/(3×4）...+1\/（48×49）+1\/(49×50）= (1 -1\/2) + (1\/2 - 1\/3) + (1\/3 - 1\/4) + ... + (1\/48 - 1\/49 + (1\/49 - 1\/50)= 1 - 1\/50 = 49\/50
1\/(1*2)+1\/(2*3)+1\/(3*4)+1\/(4*5)的简便计算的过程
原式=1-1\/2+1\/2-1\/3……即1\/(n*(n+1))=1\/n-1\/(n+1)所以原式=1-1\/5=4\/5.
1\/(1*2)+1\/(1*2*3)+1\/(1*2*3*4)+...+1\/(1*2*3*4*...*50)怎么做?有简便...
用裂项法。1\/(1*2)=1-1\/2.最后等于49\/50
计算:1\/(1*2)+1\/(2*3)+1\/(3*4)+1\/(4*5)+……1\/(99*100)=?
裂项相消：原式=1-（1\/2)+(1\/2)-(1\/3)+(1\/3)-(1\/4)+...+(1\/99)-(1\/100)=99\/100
1\/1*2+1\/2*3+1\/3*4+1\/4*5(等于多少?)我要过程!
1\/1*2+1\/2*3+1\/3*4+1\/4*5 =1-1\/2+1\/2-1\/3+1\/3-1\/4+1\/4-1\/5 =1-1\/5 =4\/5 ~~OvO~~~谢谢~！如果还有不理解的尽管追问我好了，我会乐意回答的！！祝楼主中秋快乐！>o<~!
...编写程序,求S=1\/(1*2)+1\/(2*3)+1\/(3*4)+……前50项之和。
include<stdio.h> int main(){int i;float y=0;for(i=1;i<=50;i+=2)y+=1.0\/(i*(i+1));printf("%g\\n",y);return 0;}
2.计算1\/(1*2*3)+1\/(2*3*4)+1\/(3*4*5?
运用公因数，进行通分，然后相加，得到结果。
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
1\/1*2+1\/2*3+1\/3*4+...1\/n(n+1)
==n\/n+1。1、可以分析数列的规律：1\/1×2=1-1\/2，1\/2×3=1\/2-1\/3；即每个数字都可以进行拆分为两个分数相减，通项公式为：1\/n(n+1)=1\/n-1\/n+1 2、1\/1×2+1\/2×3+1\/3×4+...1\/n(n+1)=1-1\/2+1\/2-1\/3+1\/3-1\/4+1\/n-1\/n+1=1-1\/n+1=n\/n+1。
1\/(1*2)+1\/(2*3)+1\/(3*4)+1\/(4*5)+1\/(5*6)=?
先找分母的最小公倍数，1~6的最小公倍数是60，则题目为30\/60+10\/60+5\/60+3\/60+2\/60=5\/6
