1+(1/1+2)+(1/1+2+3)+(1/1+2+3+4)`````+(1/1+2+3+`````+n)=?

来源：百度知道编辑：UC知道时间：2024/05/07 19:22:14

这里用到的裂项相消法
因为1+2+3+..+n=n(n+1)/2
所以[1/（1+2+3+…+n)]=2/n(n+1)=2[1/n-1/(n+1)]
所以Sn=1+[1/(1+2)]+〔1/(1+2+3)〕+[1/(1+2+3+4)]+……+[1/(1+2+3+……+n)]
=2[1/1-1/2]+2[1/2-1/3]+2[1/3-1/4]+...+2[1/n-1/(n+1)]
=2[1-1/2+1/2-1/3+1/3-1/4+...+1/(n-1)-1/n+1/n-1/(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=?