Question

已知数列{an}满足a1=1,an=a（n-1）/3a（n-1）+1(n>=2,n属于N*)

（1）求出an的通项公式
（2）设bn=an*a（n+1）求数列{bn}的前n项和sn
（3）求sn的极限

Answer 1

1. An=[A(n-1)]/[3A(n-1)+1]
==> 1/An =3 +1/A(n-1)
1/An -1/A(n-1) =3
==>
{1/an}为等差数列, 首项 =1/A1 =1, 公差 =3

2. 1/An =1/A1 +3(n-1) =3n-2
==> An
=1/(3n-2)

3. 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的极限为1/3

Answer 2