问题
解答题
已知数列{an}的前n项和为Sn=n2+1,数列{bn}满足:bn=
(1)求数列{bn}的通项公式; (2)是否存在自然数k,当n≥k时,总有Cn<
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答案
(1)a1=2,当n>1时,an=Sn-Sn-1=2n-1
∴bn=
,(n=1)2 3
,(n>1)1 n
(2)Cn=T2n+1-Tn=bn+1+bn+2+…+b2n+1
∵Cn+1-Cn=
+1 2n+2
-1 2n+3
=1 n+1
-1 2n+3
<01 2n+2
∴数列{Cn}是单调递减数列.
由(2)知:Cn<Cn-1<…<C3<C2<C1
当n=1时,C1=
+1 2
=1 3
>5 6 16 21
当n=2时,C2=
+1 3
+1 4
=1 5
>47 60 16 21
当n=3时,C3=
+1 4
+1 5
+1 6
=1 7
<319 420
=320 420 16 21
当n≥3时,Cn≤C3<16 21
故,kmin=3.