已知正项数列{an}满足a1=P(0<P<1),且an+1=an1+ann∈N*(1)若bn=1an,求证:数列{bn}为等差数列;(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=an1+ann∈N*(1)若bn=1an,求证:数列{bn}为等差数列;(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=P(0<P<1),且an+1=
an
1+an
n∈N*
(1)若bn=
1
an
,求证:数列{bn}为等差数列;
(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)b1=class="stub"1
a1
=class="stub"1
P

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

故数列{bn}是以b1=class="stub"1
P
为首项,以1为等差的等差数列              
(2)证明:bn=class="stub"1
an
=class="stub"1
p
+(n-1)⇒an=class="stub"1
n+class="stub"1
p
-1

0<p<1
 ∴class="stub"1
p
-1>0
an=class="stub"1
n+class="stub"1
p
-1
<class="stub"1
n

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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