已知数列{an}满足:a1=1,an-an-1+2anan-1=0,(n∈N*,n>1)(Ⅰ)求证数列{1an}是等差数列并求{an}的通项公式;(Ⅱ)设bn=anan+1,求证:b1+b2+…+bn

题目简介

已知数列{an}满足:a1=1,an-an-1+2anan-1=0,(n∈N*,n>1)(Ⅰ)求证数列{1an}是等差数列并求{an}的通项公式;(Ⅱ)设bn=anan+1,求证:b1+b2+…+bn

题目详情

已知数列{an}满足:a1=1,an-an-1+2anan-1=0,(n∈N*,n>1)
(Ⅰ)求证数列{
1
an
}
是等差数列并求{an}的通项公式;
(Ⅱ)设bn=anan+1,求证:b1+b2+…+bn
1
2
题型:解答题难度:中档来源:不详

答案

证明:(Ⅰ)an-an-1+2anan-1=0两边同除以anan-1得:class="stub"1
an
-class="stub"1
an-1
=2

所以数列{class="stub"1
an
}
是以1为首项,2为公差的等差数列…(3分)
于是class="stub"1
an
=2n-1
an=class="stub"1
2n-1
,(n∈N*)
…(6分)
(Ⅱ)由(Ⅰ),bn=class="stub"1
(2n-1)(2n+1)

b1+b2+…+bn=class="stub"1
1×3
+class="stub"1
3×5
+…+class="stub"1
(2n-1)(2n+1)

=class="stub"1
2
(1-class="stub"1
3
+class="stub"1
3
-class="stub"1
5
+…+class="stub"1
2n-1
-class="stub"1
2n+1
)
=class="stub"1
2
(1-class="stub"1
2n+1
)<class="stub"1
2
…(12分)

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