设f(x)=xa(x+2),方程f(x)=x有唯一解,数列{xn}满足f(x1)=1,xn+1=f(xn)(n∈N*).(1)求数列{xn}的通项公式;(2)已知数列{an}满足a1=12,an+1=

题目简介

设f(x)=xa(x+2),方程f(x)=x有唯一解,数列{xn}满足f(x1)=1,xn+1=f(xn)(n∈N*).(1)求数列{xn}的通项公式;(2)已知数列{an}满足a1=12,an+1=

题目详情

设f(x)=
x
a(x+2)
,方程f (x)=x有唯一解,数列{xn}满足f (x1)=1,xn+1=f (xn)(n∈N*).
(1)求数列{xn}的通项公式;
(2)已知数列{an}满足a1=
1
2
,an+1=
1
4
(2+an2-
2an
an+2
(n∈N*),求证:对一切n≥2的正整数都满足
3
4
1
x1+a1
+
1
2x2+a2
+…+
1
nxn+an
<2.
题型:解答题难度:中档来源:不详

答案

(1)由f(x)=x得ax2+(2a-1)x=0(a≠0)
∴当且仅当a=class="stub"1
2
时,f(x)=x有唯一解x=0,
f(x)=class="stub"2x
x+2

当f(x1)=
2x1
2+x1
=1得x1=2,由xn+1=f (xn)=
2xn
xn+2
可得class="stub"1
xn+1
-class="stub"1
xn
=class="stub"1
2

∴数列{class="stub"1
xn
}是首项为class="stub"1
x1
=class="stub"1
2
,公差为class="stub"1
2
的等差数列
class="stub"1
xn
=class="stub"1
2
+class="stub"1
2
(n-1)=class="stub"1
2
n

xn=class="stub"2
n

(2)∵a1=class="stub"1
2
,an+1=class="stub"1
4
(2+an)2•
2an
2+an
=
(2+an)an
2
 又a1=class="stub"1
2

class="stub"1
an+1
=class="stub"2
an(2+an)
=class="stub"1
an
-class="stub"1
an+2
 且an>0,
class="stub"1
an+2
=class="stub"1
an
-class="stub"1
an+1

class="stub"1
nxn
=class="stub"1
an
-class="stub"1
an+1

当n≥2时,class="stub"1
x1+a1
+class="stub"1
2x2+a2
+…+class="stub"1
nxn+an
≥class="stub"1
2+class="stub"1
2
+class="stub"1
2+class="stub"5
8
=class="stub"82
105
>class="stub"3
4

class="stub"1
x1+a1
+class="stub"1
2x2+a2
+…+class="stub"1
nxn+an

=(class="stub"1
a1
-class="stub"1
a2
)+(class="stub"1
a2
-class="stub"1
a3
)+…+(class="stub"1
an
-class="stub"1
an+1

=class="stub"1
a1
-class="stub"1
an+1
=2-class="stub"1
an+1
<2
∴对一切n≥2的正整数都满足class="stub"3
4
<class="stub"1
x1+a1
+class="stub"1
2x2+a2
+…+class="stub"1
nxn+an
<2.

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