解题思路:(Ⅰ)由题意知a1=a,Sn=a(Sn-an+1),Sn-1=a(Sn-1-an-1+1),由此可知an=a•an-1,
a
n
a
n-1
=a
,所以an=a•an-1=an.
(Ⅱ)由题意知a≠1,
b
n
=(
a
n
)
2
+
a(
a
n
-1)
a-1
a
n
,
b
n
=
(2a-1)
a
2n
-a
a
n
a-1
,由此可解得
a=
1
2
.
(Ⅲ)证明:由题意知
b
n
=(
1
2
)
n
,所以
c
n
=
1
(
1
2
)
n
+1
-
1
(
1
2
)
n+1
-1
=
2-
1
2
n
+1
+
1
2
n+1
-1
,由此可知Tn>2n-[1/2].
(Ⅰ)S1=a(S1-a1+1)
∴a1=a,.(1分)
当n≥2时,Sn=a(Sn-an+1),Sn-1=a(Sn-1-an-1+1),
两式相减得:an=a•an-1,
an
an−1=a
(a≠0,n≥2)即{an}是等比数列.
∴an=a•an-1=an;(4分)
(Ⅱ)由(Ⅰ)知a≠1,
bn=(an)2+
a(an−1)
a−1an,bn=
(2a−1)a2n−aan
a−1,
若{bn}为等比数列,则有b22=b1b3,
而b1=2a2,b2=a3(2a+1),b3=a4(2a2+a+1)(6分)
故[a3(2a+1)]2=2a2•a4(2a2+a+1),解得a=
1
2,(7分)
再将a=[1/2]代入得bn=([1/2])n成立,所以a=[1/2].(8分)
(Ⅲ)证明:由(Ⅱ)知bn=(
1
2)n,
所以cn=
1
(
1
2)n+1−
1
(
1
2)n+1−1=
2n
2n+1+
2n+1
2n+1−1=2−
1
2n+1+
1
2n+1−1(10分)
所以cn>2−
1
2n+
1
2n+1
Tn=c1+c2++cn>(2−
点评:
本题考点: 数列的应用;数列的求和;数列递推式.
考点点评: 本题考查数列知识的综合应用,解题时要认真审题,仔细解答.