Question

已知數(shù)列{a_n}滿足條件：a₁=1，a₂=r(r＞0)且{a_na_n+1}是公比為q(q＞0)的等比數(shù)列.設b_n=a_2n-1+a_2n(n=1，2，…).

(1)求出使不等式a_na_n+1+a_n+1a_n+2＞a_n+2a_n+3(n∈N)成立的q的取值范圍；

(2)求，其中S_n=b₁+b₂+…+b_n.

 

Answer 1

解:(1)∵{a_na_n+1}是公比為q的等比數(shù)列，且a₁=1，a₂=r,

∴a_na_n+1=rq^n－1.

∵a_na_n+1＋a_n+1a_n+2＞a_n+2a_n+3,

∴rq^n－1＋rqⁿ＞rqⁿ⁺¹.

∵r＞0，q＞0,∴1＋q＞q².

解得＜q＜.

∴0＜q＜.

(2)∵a_2n－1a_2n=rq^2n－2,∴a_2n=.                       ①

∵a_2n－1a_2n－2= rq^2n－3,

∴a_2n－1=.                                                        ②

由①②可得a_2n=qa_2n－2.                                                ③

同理a_2n－1=qa_2n－3.                                                     ④

∴b_n=a_2n－1+a_2n

=qa_2n－3＋qa_2n－2

=q(a_2n－3＋a_2n－2)

=qb_n－1.

∴{b_n}是公比為q的等比數(shù)列.

∴b_n=(a₁＋a₂)q^n－1= (1＋r)q^n－1.

∴S_n=b₁＋b₂＋…＋b_n

=(1＋r)＋(1＋r) q＋…＋(1＋r)q^n－1

=(1＋r) (1＋q＋…＋q^n－1).

當0＜q＜1時,;

當q=1時,==0;

當q＞1時,S_n=(1+r)(1+q+…+q^n-1)=(1+r),

===0.