数列{an}的前n项和为Sn,若a1=3,点(Sn,Sn+1)在直线y=n+1nx+n+1(n∈N*)上.(Ⅰ)求证:数列{Snn}是等差数列;(Ⅱ)若数列{bn}满足bn=an•2an,求数列{bn

题目简介

数列{an}的前n项和为Sn,若a1=3,点(Sn,Sn+1)在直线y=n+1nx+n+1(n∈N*)上.(Ⅰ)求证:数列{Snn}是等差数列;(Ⅱ)若数列{bn}满足bn=an•2an,求数列{bn

题目详情

数列{an}的前n项和为Sn,若a1=3,点(Sn,Sn+1)在直线y=
n+1
n
x+n+1
(n∈N*)上.
(Ⅰ)求证:数列{
Sn
n
}
是等差数列;
(Ⅱ)若数列{bn}满足bn=an2an,求数列{bn}的前n项和Tn
(Ⅲ)设Cn=
Tn
22n+3
,求证:C1+C2+…+Cn
20
27
题型:解答题难度:中档来源:宣武区一模

答案

(Ⅰ)∵点(Sn,Sn+1)在直线y=class="stub"n+1
n
x+n+1
(n∈N*)上,
Sn+1=class="stub"n+1
n
Sn+n+1

同除以n+1,则有:
Sn+1
n+1
-
Sn
n
=1

∴数列{
Sn
n
}是以3为首项,1为公差的等差数列.
(Ⅱ)由(Ⅰ)可知,Sn=n2+2n(n∈N*),∴当n=1时,a1=3,
当n≥2时,an=Sn-Sn-1=2n+1,经检验,当n=1时也成立,
∴an=2n+1(n∈N*).
bn=an2an,∴bn=(2n+1)•22n+1,
Tn=3•23+5•25++(2n-1)•22n-1+(2n+1)•22n+14Tn
=3•25++(2n-3)22n-1+(2n-1)22n+1+(2n+1)22n+3
解得:Tn=(class="stub"2
3
n+class="stub"1
9
)•22n+3-class="stub"8
9

(Ⅲ)∵Cn=
Tn
22n+3
=class="stub"2n
3
+class="stub"1
9
-class="stub"1
9
•(class="stub"1
4
)n

C1+C2+…+Cn=class="stub"2
3
n(n+1)
2
+class="stub"1
9
•n-class="stub"1
9
class="stub"1
4
[1-(class="stub"1
4
)
n
]
1-class="stub"1
4
=
3n2+4n
9
-class="stub"1
27
+class="stub"1
27
(class="stub"1
4
)
n
3n2+4n
9
-class="stub"1
27

≥class="stub"7
9
-class="stub"1
27
=class="stub"20
27

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