已知正项数列{an}满足a1=12,且an+1=an1+an(1)证明数列{1an}为等差数列,并求{an}的通项公式;(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=12,且an+1=an1+an(1)证明数列{1an}为等差数列,并求{an}的通项公式;(2)求证:a12+a23+a34+…+ann+1<1.-数学

题目详情

已知正项数列{an}满足a1=
1
2
,且an+1=
an
1+an

(1)证明数列{
1
an
}为等差数列,并求{an}的通项公式;
(2)求证:
a1
2
+
a2
3
+
a3
4
+…+
an
n+1
<1
题型:解答题难度:中档来源:不详

答案

(1)由已知得an+1an=an-an+1an
两边同除以an+1an得出
class="stub"1 
an+1
-class="stub"1
an
=1,
∴数列{class="stub"1
an
}为公差为1的等差数列,且首项为class="stub"1
a1
=2
根据等差数列的通项公式可得
class="stub"1
an
=2+(n-1)=n+1
an=class="stub"1
n+1

(2)证明:∵
an
n+1
=class="stub"1
(n+1)2
<class="stub"1
n
-class="stub"1
n+1

 
a1
2
+
a2
3
+
a3
4
+…+
an
n+1
<class="stub"1
2×1
+class="stub"1
3×2
+class="stub"1
4×3
+…+class="stub"1
(n+1)n
=1-class="stub"1
2
+class="stub"1
2
-class="stub"1
3
+class="stub"1
3
-class="stub"1
4
+…+class="stub"1
n
-class="stub"1
n+1
=1-class="stub"1
n+1
<1

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