答：
bn=1/(an^2-1)
=1/[(2n+1)^2-1]
=1/(4n^2+4n)
=1/[4n(n+1)]
=1/4[1/n-1/(1+n)]
所以Tn=b1+b2+...+bn
=1/4[1/1-1/2+1/2-1/3+...+1/n-1/(1+n)]
=1/4[1-1/(1+n)]
=n/(4+4n)