Question

已知各項(xiàng)均為正數(shù)的數(shù)列{an}的前n項(xiàng)和為Sn,滿(mǎn)足8Sn=+4an+3(n∈N*),且a1,a2,a7依次是等比數(shù)列{bn}的前三項(xiàng).

(1) 求數(shù)列{an}及{bn}的通項(xiàng)公式;

(2) 是否存在常數(shù)a>0且a≠1,使得數(shù)列{an-logabn}(n∈N*)是常數(shù)列?若存在,求出a的值;若不存在,說(shuō)明理由.

Answer 1

 (1) 當(dāng)n=1時(shí),8a1=+4a1+3,a1=1或a1=3.

當(dāng)n≥2時(shí),8Sn-1=+4an-1+3,an=Sn-Sn-1=(+4an--4an-1),

從而(an+an-1)(an-an-1-4)=0.

因?yàn)閧an}各項(xiàng)均為正數(shù),所以an-an-1=4.

所以,當(dāng)a1=1時(shí),an=4n-3;當(dāng)a1=3時(shí),an=4n-1.

又因?yàn)楫?dāng)a1=1時(shí),a1,a2,a7分別為1,5,25,構(gòu)成等比數(shù)列,

所以an=4n-3,bn=5n-1.

當(dāng)a1=3時(shí),a1,a2,a7分別為3,7,27,不構(gòu)成等比數(shù)列,舍去.綜上,an=4n-3,bn=5n-1.

(2) 由(1)知,an=4n-3,bn=5n-1,從而

an-logabn=4n-3-loga5n-1=4n-3-(n-1)loga5=(4-loga5)n-3+loga5.

由題意,得4-loga5=0,所以a=.所以,滿(mǎn)足條件的a存在,a=.