已知等差数列｛an｝满足：a3=7,a5+a7=26.｛an｝的前n项和为Sn.求令bn=1/(an)^2-1,求｛bn｝及前n项和Tn

a3=7
a5+a7=2a6=26
a6=13
a6-a3=6=5d-2d=3d,d=2
a1+2d=7=a1+4 a1=3
an=3+2(n-1)=2n+1
bn=1/[an^2-1]=1/[4n(n+1)]=(1/4)(1/n-1/(n+1))
b1=(1/4)(1-1/2)=1/8
Tn=(1/4)(1-1/(n+1))Tn=(1/4)(1-1/(n+1)) 是什么啊？b1= (1/4)(1-1/2)b2=(1/4)(1/2-1/3)..bn=(1/4)(1/n-1/(n+1))左和 =右和Sn=(1/4)(1-1/(n+1))