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.