设函数f(x)=(12)x,数列{an}满足a1=f(0),f(an+1)=1f(-2-an)(n∈N*)(1)求数列{an}的通项公式;(2)令bn=(12)an,Sn=b1+b2+…+bn,Tn=

题目简介

设函数f(x)=(12)x,数列{an}满足a1=f(0),f(an+1)=1f(-2-an)(n∈N*)(1)求数列{an}的通项公式;(2)令bn=(12)an,Sn=b1+b2+…+bn,Tn=

题目详情

设函数f(x)=(
1
2
)x
,数列{an}满足a1=f(0),f(an+1)=
1
f(-2-an)
(n∈N*)

(1)求数列{an}的通项公式;
(2)令 bn=(
1
2
)anSn=b1+b2+…+bnTn=
1
a1a2
+
1
a2a3
+…+
1
anan+1
,试比较 Sn
4
3
Tn
的大小,并加以证明.
题型:解答题难度:中档来源:不详

答案

(1)∵f(x)=(class="stub"1
2
)xa1=f(0)=(class="stub"1
2
)0=1

又∵f(an+1)=class="stub"1
f(-2-an)

(class="stub"1
2
)an+1=class="stub"1
(class="stub"1
2
)
-2-an
=(class="stub"1
2
)an+2
.…(2分)
∴an+1=an+2即 an+1-an=2,∴数列{an}是首项为1,公差为 2 的等差数列
∴an=1+(n-1)×2=2n-1.…(5分)
(2)∵bn=(class="stub"1
2
)an=(class="stub"1
2
)2n-1
bn+1
bn
=
(class="stub"1
2
)
2n+1
(class="stub"1
2
)
2n-1
=class="stub"1
4
…(6分)
即数列{bn}是首项为 class="stub"1
2
,公比为 class="stub"1
4
的等比数列
Sn=b1+b2+…+bn=
class="stub"1
2
[1-(class="stub"1
4
)
n
]
1-class="stub"1
4
=class="stub"2
3
[1-(class="stub"1
4
)n]
…(7分)Tn=class="stub"1
a1a2
+class="stub"1
a2a3
+…+class="stub"1
anan-1
=class="stub"1
1×3
+class="stub"1
3×5
+…+class="stub"1
(2n-1)(2n+1)
=class="stub"1
2
[(1-class="stub"1
3
)+(class="stub"1
3
-class="stub"1
5
)+…+(class="stub"1
2n-1
-class="stub"1
2n+1
)]=class="stub"1
2
(1-class="stub"1
2n+1
)
…(10分)
class="stub"4
3
Tn=class="stub"2
3
(1-class="stub"1
2n+1
)

故比较Sn与class="stub"4
3
Tn
的大小,只需比较 (class="stub"1
4
)n
class="stub"1
2n+1
的大小即可       …(11分)
即只需比较 2n+1与4n的大小
∵4n=(1+3)n=1+Cn1•3+…≥3n+1>2n+1…(12分)
故 Sn>class="stub"4
3
Tn
   …(13分)

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