已知数列an=1/n(n+1)(n+2),求数列的前n项和Sn 最好利用裂项法

题目

已知数列an=1/n(n+1)(n+2),求数列的前n项和Sn 最好利用裂项法

答案

an=1/2*2/[n(n+1)(n+2)]=1/2*[(n+2)-n]/[n(n+1)(n+2)]=1/2{[(n+2)/[n(n+1)(n+2)]-n/[n(n+1)(n+2)]=1/2{1/[n(n+1)]-1/[(n+1)(n+2)]}所以Sn=1/2*{1/1*2-1/2*3+1/2*3-1/3*4+……+1/[n(n+1)]-1/[(n+1)(n+2)]}=1/2*{1/1*2...