Question

设数列{a_n}的前n项和为S_n,且(3-m)S_n+2ma_n=m+3(n∈N),其中m为常数,且m≠-3.

(1)求证:{a_n}是等比数列;

(2)若数列{a_n}的公比q=f(m),数列{b_n}满足b₁=a₁,b_n=f(b_n-1)(n∈N,n≥2),求证:{}为等差数列.

Answer 1

思路分析:本题要求证明数列为等差、等比数列,分析的思路是定义,所以恰当地处理递推关系是关键.

证明:(1)由(3-m)S_n+2ma_n=m+3,得(3-m)S_n+1+2ma_n+1=m+3,

两式相减得(3+m)a_n+1=2ma_n,m≠-3,

∴.∴{a_n}是等比数列.

(2)b₁=a₁=1,q=f(m)=,

∴n∈N,n≥2时,b_n=f(b_n-1)=·b_nb_n-1+3b_n=3b_n-1=.

∴{}是首项为1,公差为的等差数列.