看通项an=[(n+1)²+1]/[(n+1)²-1]
=1+2/[(n+1)²-1]
=1+2/[(n+1-1)*(n+1+1)]
=1+2/[n(n+2)]
=1+1/n-1/(n+2)
∴ 前n项和=(1+1/1-1/3)+(1+1/2-1/4)+(1-1/3-1/5)+.+[1+1/n-1/(n+2)]
=n+1-1/3+1/2-1/4+1/3-1/5+.+1/(n-1)-1/(n+1)+1/n-1/(n+2)
=n+1+1/2-1/(n+1)-1/(n+2)