高二数列有道解题看不懂
已知数列a1=1,an=a(n-1)/3a(n-1)+1(n>=2)设bn=ana(n+1),求数列{an}的通项公式求数列{bn}的前n项和sn
An=[A(n-1)]/[3A(n-1)+1]
==>1/An=3+1/A(n-1)
==>{1/an}为等差数列,首项=1/A1=1,公差=3
1/An=1/A1+3(n-1)=3n-2
==>An=1/(3n-2)
Bn=An*A(n+1)=1/(3n-2)(3n+1)=[1/(3n-2)-1/(3n+1)]/3
==>Sn=[1-1/(3n+1)]/3=n/(3n+1)
Sn=[1-1/(3n+1)]/3=n/(3n+1)是怎么得出来的,没有公式可循啊.