1/(1+2)+1/(1+2+3)+1/(1+2+3+4)......1/(1+2+3+......+2001)

来源：百度知道编辑：UC知道时间：2024/06/23 01:11:14

不能只是答案，还要有公式。

裂项相消法
数列的通项为2/（n(n+1))=2/n-2/(n+1)
1/(1+2)+1/(1+2+3)+1/(1+2+3+4)......2/(n(n+1))=2/2-2/(n+1)
1/(1+2)+1/(1+2+3)+1/(1+2+3+4)......1/(1+2+3+......+2001)
=2/2-2/2001=1-2/2001=1999/2001

写出通项:a1=1
a2=1/(1+2)
....
an = 1/(1+2+....n)=1/[n(n+1)/2]=2/[n(n+1)=2[1/n-1/(n+1)]
所以a(n-1)=2[1/(n-1) -1/n] ;每一项与前一项都有一部分可以约去. 则原式=1+2[1/2-1/3]+........+2[1/(n-1)-1/n]+2[1/n-1/(n+1)]=1+2*1/2 - 2/(n+1)= 2 - 2/(n+1) =2 [1-1/(n+1)]=2n/(n+1).

(1/2005-1)(1/2004-1)........(1/3-1)(1/2-1) 1+1/(1+2)+1/(1+2+3)+...+1/(1+2+3+...+100) 1+1/(1+2)+1/(1+2+3)+-------+1/(1+2+3+----+100) 1+1/1+2+1/1+2+3+...+1/1+2+3...+2000 1+1/1+2+1/1+2+3.........+1/1+2+3.....100 1*(1/1+2)*(1/1+2+3)*~~~*(1/1+2+~~~2005)=? (1-1/2)(1-1/3)(1-1/4)(1-1/5).....(1-1/1000) (1+1/2)(1+1/2^2)(1+1/2^4)(1+1/2^8) (1-1/2^2)*(1-1/3^2)*(1-1/4^2).......(1-1/100^2) 1+1/2+1+1/3+1+1/4+......+1/100=?