问题
解答题
设{a}是正数数列,其前n项和Sn满足Sn=
(1)求a1的值;求数列{an}的通项公式; (2)对于数列{bn},令bn=
|
答案
(1)由a1=S1=
(a1-1)(a1+3),及an>0,得a1=31 4
由Sn=
(an-1)(an+3)得Sn-1=1 4
(an-1-1)(an-1+3).1 4
∴当n≥2时,an=
(1 4
-a 2n
)+2(an-an-1)a 2n-1
∴2(an+an-1)=(an+an-1)(an-an-1)∵an+an-1>0∴an-an-1=2,
∴{an}是以3为首项,2为公差的等差数列,∴an=2n+1
(2)由(1)知Sn=n(n+2)∴bn=
=1 Sn
(1 2
-1 n
),1 n+2
Tn=b1+b2+…+bn
=
(1-1 2
+1 3
-1 2
++1 4
-1 n-1
+1 n+1
-1 n
)1 n+2
=
[1 2
-3 2
]=2n+3 (n+1)(n+2)
-3 4 2n+3 2(n+1)(n+2)
∴
Tn=lim n→∞
[lim n→∞
-3 4
]=2n+3 2(n+1)(n+2)
(13分)3 4
由
<0,得a1+(b1-a1)•(an+bn 2
)n<01 2
得
<-a,得a1+b1 2n
<2nb1-a1 -a1
∴log2
<na1-b1 a1
因而n满足log2
<n的最小整数(14分)a1-b1 a1