Question

1\1+2+1\1+2+3+1\1+2+3+4+......1\1+2...20

Answer 1

http://zhidao.baidu.com/link?url=suzOQW3ml8I0Q_wMnEzPR5kHHZECzdXOXdvexUDnptf9uiHit9aKSWHdN-4ox_aycnI9X8dI7ux5P0OQdlONoa
因为1+2+3+ …+n=n(n+1)/2
所以1/(1+2+3+ …+ n)=2[1/n-1/(n+1)]
所以1+1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+ …+1/(1+2+3+ …+n)
=1+2[1/2-1/3+1/3-1/4+1/4-1/5+1/n-1/(n+1)]
=1+2×[1/2-1/(n+1)]
=1+1-2/(n+1)
=2-2/(n+1)
=2n/(n+1)
将20带入，再减去1，等于19/21.

追问

能运用裂项吗

追答

裂项下面是相乘的。。最简单的裂项法就是1/[n(n+1)]=（1/n）- [1/(n+1)]