求数列 1除以1+2 加 1除以1+2+3 ……加1除以1+2+……+（n+1）的前n项和

网友回答

S(N)=1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+.+1/[1+2+.+(n+1)]
S(n)=(1+2+.+(n+1)=[1+(n+1)]*n/2=[(n+2)*(n+1)]/2
1/[S(n)]=2/[(n+2)*(n+1)]
=2*{1/[(n+2)*(n+1)]}
=2*[1/(n+1)-1/(n+2)]
代入“S(N)=1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+.+1/[1+2+.+(n+1)]”中得：
S(N)=2*[1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+.+1/(n+1)-1/(n+2)]
=2*[1/2-1/(n+2)]
=1-2*[1/(n+2)]
=1-2/(n+2)