解题思路:(Ⅰ)由题设条件能导出an+1-an=5an+1,即
a
n+1
=-
1
4
a
n
,所以
a
n
=(-
1
4
)
n
,∴
b
n
=
4+
(-
1
4
)
n
1-
(-
1
4
)
n
.
(Ⅱ)由
b
n
=4+
5
(-4)
n
-1
,知
c
n
=
b
2n
-
b
2n-1
=
5
4
2n
-1
+
5
4
2n-1
+1
=
25×
16
n
(
16
n
-1)(
16
n
+4)
=
25×
16
n
(
16
n
)
2
+3×
16
n
-4
<
25×
16
n
(
16
n
)
2
=
25
16
n
,当n=1时,
T
1
<
3
2
;当n≥2时,
T
n
<
4
3
+25×(
1
16
2
+
1
16
3
+…+
1
16
n
)
<
4
3
+25×
1
16
2
1-
1
16
=
69
48
<
3
2
.
(Ⅲ)由
b
n
=4+
5
(-4)
n
-1
知Rn=b1+b2+…+b2k+1=
4n+5×(-
1
4
1
+1
+
1
4
2
-1
-
1
4
3
+1
+…-
1
4
2k+1
+1
)
=
4n+5×[-
1
4
1
+1
+(
1
4
2
-1
-
1
4
3
+1
)+…+(
1
4
2k
-1
-
1
4
2k+1
+1
)]
>4n-1.由此入手能推导出正实数λ的最小值为4.
(Ⅰ)当n=1时,a1=5a1+1,∴a1=−
1
4
又∵an=5Sn+1,an+1=5Sn+1+1
∴an+1-an=5an+1,即an+1=−
1
4an
∴数列an成等比数列,其首项a1=−
1
4,公比是q=−
1
4
∴an=(−
1
4)n
∴bn=
4+(−
1
4)n
1−(−
1
4)n
(Ⅱ)由(Ⅰ)知bn=4+
5
(−4)n−1
∴cn=b2n−b2n−1=
5
42n−1+
5
42n−1+1=
25×16n
(16n−1)(16n+4)
=
25×16n
(16n)2+3×16n−4<
25×16n
(16n)2=
25
16n
又b1=3,b2=
13
3,∴c1=
4
3
当n=1时,T1<
3
2
当n≥2时,Tn<
4
3+25×(
点评:
本题考点: 数列递推式;数列的求和;数列与不等式的综合.
考点点评: 本题主要考查数列、不等式等基础知识、考查化归思想、分类整合思想,以及推理论证、分析与解决问题的能力.