已知函数f(x)=2xx+1(1)当x≥1时,证明:不等式f(x)≤x+lnx恒成立.(2)若数列{an}满足a1=23,an+1=f(an),bn=1an-1,n∈N+,证明数列{bn}是等比数列,

题目简介

已知函数f(x)=2xx+1(1)当x≥1时,证明:不等式f(x)≤x+lnx恒成立.(2)若数列{an}满足a1=23,an+1=f(an),bn=1an-1,n∈N+,证明数列{bn}是等比数列,

题目详情

已知函数f(x)=
2x
x+1

(1)当x≥1时,证明:不等式f(x)≤x+lnx恒成立.
(2)若数列{an}满足a1=
2
3
,an+1=f(an),bn=
1
an
-1
,n∈N+,证明数列{bn}是等比数列,并求出数列{bn}、{an}的通项公式;
(3)在(2)的条件下,若cn=an•an+1•bn+1(n∈N+),证明:c1+c2+c3+…cn
1
3
题型:解答题难度:中档来源:不详

答案

(1)∵x≥1得f(x)-x=class="stub"2x
x+1
-x=
2x-x2-x
x+1
=
-x(x-1)
x+1
≤0,
而x≥1时,lnx≥0
∵x≥1时,f(x)-x≤lnx
∴当x≥1时,f(x)≤x+lnx恒成立
(2)a1=class="stub"2
3
,an+1=f(an),bn=class="stub"1
an
-1,n∈N+∴an+1=
2an
an+1
class="stub"1
an+1
=class="stub"1
2
+class="stub"1
2an

∴a1=class="stub"2
3
,an+1=f(an),bn=class="stub"1
an
-1,n∈N+
bn+1
bn
=
class="stub"1
an+1
-1
class="stub"1
an
-1
=
class="stub"1
2
+class="stub"1
2an
-1
class="stub"1
an
-1
=
class="stub"1
2an
-class="stub"1
2
class="stub"1
an
-1
=class="stub"1
2
(n∈N+)
又b1=class="stub"1
a1
-1=class="stub"1
2
∴{bn}是首项为class="stub"1
2
,公比为class="stub"1
2
的等比数列,其通项公式为bn=class="stub"1
2n

又a1=class="stub"2
3
,an+1=f(an),bn=class="stub"1
an
-1,n∈N+
∴an=class="stub"1
bn+1
=class="stub"1
class="stub"1
2n
+1
=
2n
2n+1
(n∈N+)
(3)cn=an•an+1•bn+1=
2n
2n+1
×
2n+1
2n+1+1
×class="stub"1
2n+1
=
2n
2n+1
×class="stub"1
2n+1+1
=class="stub"1
2n+1
-class="stub"1
2n+1+1

∴c1+c2+c3+…+cn=(class="stub"1
21+1
-class="stub"1
22+1
)+(class="stub"1
22+1
-class="stub"1
23+1
)+…+(class="stub"1
2n+1
-class="stub"1
2n+1+1
)=class="stub"1
3
-class="stub"1
2n+1+1
class="stub"1
3

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