1/(2×3)=(1/2)－(1/3)
1/(3×4)=(1/3)－(1/4)
1/(4×5)=(1/4)－(1/5)
…………
1/[n(n＋)]=[1/n]－1/[n＋1]
则：
原式=(1/1)－1/(n＋1)=n/(n＋1)
1除以1*2+1除以2*3+1除以3*4一直到1除以n(n+1)
1\/(2×3)=(1\/2)－(1\/3)1\/(3×4)=(1\/3)－(1\/4)1\/(4×5)=(1\/4)－(1\/5)………1\/[n(n＋)]=[1\/n]－1\/[n＋1]则：原式=(1\/1)－1\/(n＋1)=n\/(n＋1)
1除以1*2+1除以2*3+1除以3*4一直到1除以n(n+1)
1\/(2×3)=(1\/2)－(1\/3)1\/(3×4)=(1\/3)－(1\/4)1\/(4×5)=(1\/4)－(1\/5)………1\/[n(n＋)]=[1\/n]－1\/[n＋1]则：原式=(1\/1)－1\/(n＋1)=n\/(n＋1)
1\/1*2+1\/2*3+1\/3*4+...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\/n(n+1)
=1\/n-1\/(n+1)所以A1=1-1\/(1+1)1-1\/2 A2=1\/2-1\/3 A3=1\/3-1\/4 。。。An=1\/n-1\/(n+1)所以1\/1*2+1\/2*3+1\/3*4+…+1\/n(n+1)就等于 A1+A2+A3+...+An= 1-1\/2+1\/2-1\/3+1\/3-1\/4+。。。+1\/n-1\/(n+1)＝1-1\/(n+1)＝n\/(n+1)这考的是裂项相...
1\/1*2+1\/2*3+1\/3*4+.1\/n(n+1)
急！1\/1*2+1\/2*3+1\/3*4+.+1\/n(n+1)+? 1\/(1*2)=1-1\/2 1\/(3*4)=1\/3-1\/4 1\/[n(n+1)]=1\/n-1\/(n+1) 所以 原式=1-1\/2 +1\/2-1\/3 +1\/3-1\/4 +... +1\/n -1\/(n+1) 中间项两两相加，全得0 =1-1\/(n+1) =n\/(n+1)求证：1\/2+1\/3...
计算1\/1×2+1\/2x3+1\/3×4+...+1\/n(n+1)
y=5如此类推以后...最后1项 = 1\/n*(n+1) = (1\/n)-[1\/(n+1)] <--- 假设 x=n 且自然的 y=n+1原式 = 1\/1*2+1\/2*3+1\/3*4+1\/4*5+...+1\/n*(n+1)= (1\/1-1\/2)+(1\/2-1\/3)+(1\/3-1\/4)+(1\/4-1\/5)+...+[1\/(n-1)-1\/n]...
1\/1*2+1\/2*3+…+1\/n(n+1) 怎么做?
1\/1*2+1\/2*3+…+1\/n(n+1)=1-1\/2+1\/2-1\/3+...+1\/n-1\/(n+1)=1-1\/(n+1)=n\/(n+1);您好，很高兴为您解答，skyhunter002为您答疑解惑 如果本题有什么不明白可以追问，如果满意记得采纳 如果有其他问题请采纳本题后另发点击向我求助，答题不易，请谅解，谢谢。祝学习进步 ...
1\/1*2+1\/2*6+1\/3*4……1\/n(n+1)
中间写错一个数了 解：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)
已知,数列1\/1*2,1\/2*3,1\/3*4,...,1\/n(kn+1),求其前n项的和
根据:1\/n(n+1)=1\/n-1\/(n+1)1\/1*2+1\/2*3+1\/3*4+.1\/n(n+1)=1\/1-1\/2+1\/2-1\/3+1\/3-1\/4+.+1\/(n-1)-1\/n +1\/n-1\/(n+1)=1-1\/(n+1)=n\/(n+1)
已知一个数列1\/1X2,1\/2X3,1\/3X4,…1\/n(n+1)…,则前n项的和Sn=?
因为1\/n(n+1)=1\/n-1\/(n+1)所以 Sn=1\/1X2+1\/2X3+1\/3X4+…+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)求1\/1X3+1\/3X5+1\/5X7,…1\/（2n-1)(2n+1)也是用拆项法来解 利用1\/（2n-1)(2n+1)=(1\/2)[...
