问题
解答题
已知数列{an}的前n项和Sn和通项an满足
(Ⅰ)求数列{an}的通项公式; (Ⅱ)当q=
(Ⅲ)设函数.f(x)=logqx,bn=f(a1)+f(a2)+…+f(an),使
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答案
(I )当n≥2时,an=Sn-Sn-1=
(an-1)-q q-1
(an-1-1),∴q q-1
=q,又由S1=a1=an an-1
(a1-1)得a1=q,∴数列an是首项a1=q、公比为q的等比数列,∴an=q•qn-1=qnq q-1
(II)a1+a2+…+an=
=
[1-(1 4
)n]1 4 1- 1 4
(1-1 3
)<1 4n 1 3
(III)bn=logqa1+logqa2+…+logqan=logq(a1a2…an)=logqq1+2+n=n(n+1) 2
∴
+1 b1
++1 b2
=2(1-1 bn
+1 2
-1 2
+1 3
-1 n
),∴2(1-1 n+1
)≥1 n+1
即m≤6(1-m 3
)1 n+1
∵n=1时,[6(1-
)]min=3,∴m≤3,∵m是正整数,∴m的值为1,2,31 n+1