Question

設(shè)數(shù)列{a_n}的前n項(xiàng)和為S_n,且(3-m)S_n+2ma_n=m+3(n∈N),其中m為常數(shù),且m≠-3.

(1)求證:{a_n}是等比數(shù)列;

(2)若數(shù)列{a_n}的公比q=f(m),數(shù)列{b_n}滿足b₁=a₁,b_n=f(b_n-1)(n∈N,n≥2),求證:{}為等差數(shù)列.

Answer 1

思路分析:本題要求證明數(shù)列為等差、等比數(shù)列,分析的思路是定義,所以恰當(dāng)?shù)靥幚磉f推關(guān)系是關(guān)鍵.

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

兩式相減得(3+m)a_n+1=2ma_n,m≠-3,

∴.∴{a_n}是等比數(shù)列.

(2)b₁=a₁=1,q=f(m)=,

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

∴{}是首項(xiàng)為1,公差為的等差數(shù)列.