由t^2-2a(n-1)t+a(n-1)an=0
得t^2-a(n-1)t+a(n-1)an-a(n-1)t=0
t（t-a(n-1))+a(n-1)(an-t)=0
a(n-1)(an-t)=t(a(n-1)-t)
a(n-1)/(a(n-1)-t)=t/(an-t)
1/（an-t)-1(a(n-1)-t)=1/t
所以 数列{1/（an-t)}为以1/t为首项以1/t为公比的等差数列
1/(an-t)=n/t
an=t/n+t
bn=t/{n(n-1)}=t{1/n-1/(n+1)}
所以Sn=t{1-1/2+1/2-1/3.+1/n-1/(n+1)}=t{1-1/(n+1)}