Question

已知數(shù)列{a_n}的前n項和為S_n,點（n,a_n）(n∈N^*)在函數(shù)f(x)=-2x-2的圖象上，數(shù)列{b_n}的前n項和為T_n,且T_n是6S_n與8n的等差中項.

(1)求數(shù)列{b_n}的通項公式;

(2)設c_n=b_n+8n+3,數(shù)列{d_n}滿足d₁=c₁,d_n+1=cd_n(n∈N^*).求數(shù)列{d_n}的前n項和D_n;

(3)設g(x)是定義在正整數(shù)集上的函數(shù),對于任意的正整數(shù)x₁,x₂恒有g(x₁x₂)=x₁g(x₂)+x₂g(x₁)成立,且g(2)=a(a為常數(shù),a≠0),試判斷數(shù)列{}是否為等差數(shù)列,并說明理由.

Answer 1

解:(1)依題意得a_n=-2n-2,故a₁=-4.

又2T_n=6S_n+8n,即T_n=3S_n+4n,

所以,當n≥2時,b_n=T_n-T_n-1=3(S_n-S_n-1)+4=3a_n+4=-6n-2.

又b₁=T₁=3S₁+4=3a₁+4=-8,也適合上式,

故b_n=-6n-2(n∈N^*).

(2)因為c_n=b_n+8n+3=-6n-2+8n+3=2n+1(n∈N^*),

d_n+1==2d_n+1,因此d_n+1+1=2(d_n+1)(n∈N^*).

由于d₁=c₁=3,

所以{d_n+1}是首項為d₁+1=4,公比為2的等比數(shù)列.

故d_n+1=4×2^n-1=2ⁿ⁺¹,所以d_n=2ⁿ⁺¹-1.

所以D_n=(2²+2³+…+2ⁿ⁺¹)-n=-n=2ⁿ⁺²-n-4.

(3)方法一:g()=g(2ⁿ)=2^n-1g(2)+2g(2^n-1).

則====.

所以=.

因為a為常數(shù)，則數(shù)列{}是等差數(shù)列.

方法二:因為g(x₁x₂)=x₁g(x₂)+x₂g(x₁)成立，

且g(2)=a,故g()=g(2ⁿ)=2^n-1g(2)+2g(2^n-1)

=2^n-1g(2)+2［2^n-2g(2)+2g(2^n-2)］=2×2^n-1g(2)+2²g(2^n-2)

=2×2^n-1g(2)+2²［2^n-3g(2)+2g(2^n-3)］=3×2^n-1g(2)+2³g(2^n-3)

=…=(n-1)×2^n-1g(2)+2^n-1g(2)

=n·2^n-1g(2)=an·2^n-1,

所以==n.

因此，數(shù)列{}是等差數(shù)列.