Question

已知数列{a_n}的前n项和为S_n,点(n,a_n)(n∈N^*)在函数f(x)=-6x-2的图象上.

(1)求S_n;

(2)设c_n=a_n+8n+3,数列{d_n}满足d₁=c₁,d_n+1=(n∈N^*),求数列{d_n}的通项公式;

(3)设g(x)是定义在正整数集上的函数,对于任意的正整数x₁、x₂,恒有g(x₁x₂)=x₁g(x₂)+x₂g(x₁)成立,且g(2)=a(a为常数,且a≠0),记b_n=,试判断数列{b_n}是否为等差数列,并说明理由.

Answer 1

解:(1)由已知a_n=-6n-2,故{a_n}是以a₁=-8为首项公差为-6的等差数列.

所以S_n=-3n²-5n.

(2)因为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}是首项为d₁+1=4,公比为2的等比数列.

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),

则b_n==+,b_n+1=+.

b_n+1-b_n===.

因为a为常数,则数列{b_n}是等差数列.

方法二:因为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.

则b_n+1-b_n=.

由a为常数,因此,数列{b_n}是等差数列.