已知数列{an}的首项a1=35,an+1=3an2an+1,n=1,2,….(Ⅰ)求{an}的通项公式;(Ⅱ)证明:对任意的x>0,an≥11+x-1(1+x)2(23n-x),n=1,2,…;(Ⅲ

题目简介

已知数列{an}的首项a1=35,an+1=3an2an+1,n=1,2,….(Ⅰ)求{an}的通项公式;(Ⅱ)证明:对任意的x>0,an≥11+x-1(1+x)2(23n-x),n=1,2,…;(Ⅲ

题目详情

已知数列{an}的首项a1=
3
5
an+1=
3an
2an+1
,n=1,2,….
(Ⅰ)求{an}的通项公式;
(Ⅱ)证明:对任意的x>0,an
1
1+x
-
1
(1+x)2
(
2
3n
-x)
,n=1,2,…;
(Ⅲ)证明:a1+a2+…+an
n2
n+1
题型:解答题难度:中档来源:陕西

答案

(Ⅰ)∵an+1=
3an
2an+1
,∴class="stub"1
an+1
=class="stub"2
3
+class="stub"1
3an

class="stub"1
an+1
-1=class="stub"1
3
(class="stub"1
an
-1)

class="stub"1
an
-1=class="stub"2
3

(class="stub"1
an
-1)
是以class="stub"2
3
为首项,class="stub"1
3
为公比的等比数列.
class="stub"1
an
-1=class="stub"2
3
•class="stub"1
3n-1
=class="stub"2
3n
,∴an=
3n
3n+2

(Ⅱ)由(Ⅰ)知an=
3n
3n+2
>0
class="stub"1
1+x
-class="stub"1
(1+x)2
(class="stub"2
3n
-x)
=class="stub"1
1+x
-class="stub"1
(1+x)2
(class="stub"2
3n
+1-1-x)
=class="stub"1
1+x
-class="stub"1
(1+x)2
[class="stub"1
an
-(1+x)]
=-class="stub"1
an
•class="stub"1
(1+x)2
+class="stub"2
1+x
=-class="stub"1
an
(class="stub"1
1+x
-an)2+an
≤an,
∴原不等式成立.
(Ⅲ)由(Ⅱ)知,对任意的x>0,有a1+a2++an≥class="stub"1
1+x
-class="stub"1
(1+x)2
(class="stub"2
3
-x)+class="stub"1
1+x
-class="stub"1
(1+x)2
(class="stub"2
32
-x)++class="stub"1
1+x
-class="stub"1
(1+x)2
(class="stub"2
3n
-x)
=class="stub"n
1+x
-class="stub"1
(1+x)2
(class="stub"2
3
+class="stub"2
32
++class="stub"2
3n
-nx)

∴取x=class="stub"1
n
(class="stub"2
3
+class="stub"2
32
++class="stub"2
3n
)=
class="stub"2
3
(1-class="stub"1
3n
)
n(1-class="stub"1
3
)
=class="stub"1
n
(1-class="stub"1
3n
)

a1+a2++an≥class="stub"n
1+class="stub"1
n
(1-class="stub"1
3n
)
=
n2
n+1-class="stub"1
3n
n2
n+1
.∴原不等式成立.

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