Question

已知数列{an}满足a1=1,an=[a(n-1)]/[3a(n-1)+1]

2. 求数列{an}的通项公式；
3. 设bn=ana(n+1)，求数列{bn}的前n项和Sn 最后一问的过程

Answer 1

An=[A(n-1)]/[3A(n-1)+1]
取倒数得: 1/An =3 +1/A(n-1)
即:{1/an}为等差数列, 首项 =1/A1 =1, 公差 =3
1/An =1/A1 +3(n-1) =3n-2
所以, An =1/(3n-2)
(2)
Bn =An*A(n+1) =1/(3n-2)(3n+1) =[1/(3n-2) -1/(3n+1)]/3
Sn =1/1*4+1/4*7+...+1/(3n-2)(3n+1)
=1/3[1-1/4+1/4-1/4+...+1/(3n-2)-1/(3n+1)]
=[1 -1/(3n+1)]/3
= n/(3n+1)

Answer 2

部值得