解题思路:(Ⅰ)由已知条件得
a
n+1
+1=(
a
n
+1
)
2
,两边取对数,得lg(an+1+1)=2lg(an+1),由此能证明数列{lg(an+1)}是以1为首项,以2为公比的等比数列.
(Ⅱ)由(Ⅰ)知
lg(a
n
+1)=
2
n−1
,从而lgTn=lg(a1+1)+lg(a2+1)+…+lg(an+1),由此求出lgTn=2n-1.
(Ⅲ)由Cn=
2
n
(
2
n
−1)(
2
n+1
−1)
=
1
2
n
−1
−
1
2
n+1
−1
,利用裂项求和法能求出数列{Cn}的前n项和.
(Ⅰ)证明:∵数列{an}中,a1=9,点(an,an+1)在函数f(x)=x2+2x的图象上,
∴an+1=an2+2an,∴an+1+1=(an+1)2,
对an+1+1=(an+1)2两边取对数,得lg(an+1+1)=2lg(an+1),
∵a1=9,∴lg(a1+1)=lg10=1,
∴数列{lg(an+1)}是以1为首项,以2为公比的等比数列.
(Ⅱ)由(Ⅰ)知lg(an+1)=2n−1,
Tn=(a1+1)…(an+1),
∴lgTn=lg(a1+1)(a2+1)…(an+1)
=lg(a1+1)+lg(a2+1)+…+lg(an+1)
=20+21+22+…+2n-1
=
1−2n
1−2=2n-1.
(Ⅲ)证明:Cn=
lgTn+1
[lg(an+1+1)−1][lg(an+2+1)−1]
=
2n
(2n−1)(2n+1−1)=[1
2n−1−
1
2n+1−1,
Sn=(
1/2−1−
1
22−1])+(
1
22−1−
1
23−1)+(
1
23−1−
1
24−1)+…+(
1
2n−1−
1
2n+1−1)
=1-
1
2n+1−1<1.
∴Sn<1.
点评:
本题考点: 数列与函数的综合;数列的求和.
考点点评: 本题考查等比数列的证明,考查数列的前n项和的求法,考查不等式的证明,解题时要认真审题,注意裂项求和法的合理运用.