An=1/(n+2)n=1/2[1/n-1/(n+2)] 所以前N项和Sn=1/2[1-1/3+1/2-1/4+1/3-1/5.1/(n-2)-1/n+1/(n-1)-1/(n+1)+1/(n-1)/(n+2)]=1/2[1+1/2-1/(n+1)-1/(n+2)]=3/4-1/2(n+1)-1/2(n+2)