已知正项数列{an}满足a1=P(0<P<1),且an+1=anan+1,(1)求数列的通项an;(2)求证:a12+a23+a34+…+ann+1<1.-数学

题目简介

已知正项数列{an}满足a1=P(0<P<1),且an+1=anan+1,(1)求数列的通项an;(2)求证:a12+a23+a34+…+ann+1<1.-数学

题目详情

已知正项数列{an}满足
a 1
=P(0<P<1),且
a n+1
=
a n
a n
+1

(1)求数列的通项an
(2)求证:
a 1
2
+
a 2
3
+
a 3
4
+…+
a n
n+1
<1
题型:解答题难度:中档来源:不详

答案

由已知an+1=
an
an+1
可得,class="stub"1
an+1
=
an+1
an
=class="stub"1
an
+1
class="stub"1
a1
=class="stub"1
p

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

数列{class="stub"1
an
}是以class="stub"1
p
为首项,以1为公差的等差数列
class="stub"1
an
=class="stub"1
p
+(n-1)×1=n-1+class="stub"1
p
an=class="stub"1
n-1+class="stub"1
p

∵0<P<1∴class="stub"1
p
-1>0

an
n+1
=class="stub"1
(n+1)(n-1+class="stub"1
p
)
<class="stub"1
n(n+1)
=class="stub"1
n
-class="stub"1
n+1

a1
2
+
a2
3
+…+ 
an
n+1
 <
1-class="stub"1
2
+class="stub"1
2
-class="stub"1
3
+…+class="stub"1
n
-class="stub"1
n+1
=1-class="stub"1
n+1
=class="stub"n
n+1
<1
即证

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