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)=-6x-2的圖象上.

(1)求S_n;

(2)設c_n=a_n+8n+3,數(shù)列{d_n}滿足d₁=c₁,d_n+1=(n∈N^*),求數(shù)列{d_n}的通項公式;

(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),且a≠0),記b_n=,試判斷數(shù)列{b_n}是否為等差數(shù)列,并說明理由.

Answer 1

解:(1)由已知a_n=-6n-2,故{a_n}是以a₁=-8為首項公差為-6的等差數(shù)列.

所以S_n=-3n²-5n.

(2)因為c_n=a_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.

(3)方法一:g()=g(2ⁿ)=2^n-1g(2)+2g(2^n-1),

則b_n==+,b_n+1=+.

b_n+1-b_n===.

因為a為常數(shù),則數(shù)列{b_n}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b_n= n.

則b_n+1-b_n=.

由a為常數(shù),因此,數(shù)列{b_n}是等差數(shù)列.