数列{an}满足:a1=14,a2=15,且a1a2+a2a3+…+anan+1=na1an+1对任何的正整数n都成立,则1a1+1a2+…+1a97的值为()A.5032B.5044C.5048D.

题目简介

数列{an}满足:a1=14,a2=15,且a1a2+a2a3+…+anan+1=na1an+1对任何的正整数n都成立,则1a1+1a2+…+1a97的值为()A.5032B.5044C.5048D.

题目详情

数列{an}满足:a1=
1
4
,a2=
1
5
,且a1a2+a2a3+…+anan+1=na1an+1对任何的正整数n都成立,则
1
a1
+
1
a2
+…+
1
a97
的值为(  )
A.5032B.5044C.5048D.5050
题型:单选题难度:偏易来源:不详

答案

a1a2+a2a3+…+anan+1=na1an+1,①
a1a2+a2a3+…+anan+1+an+1an+2=(n+1)a1an+2,②
①-②,得-an+1an+2=na1an+1-(n+1)a1an+2,
class="stub"n+1
an+1
-class="stub"n
an+2
=4

同理,得class="stub"n
an
-class="stub"n-1
an+1
=4,
class="stub"n+1
an+1
-class="stub"n
an+2
=class="stub"n
an
-class="stub"n-1
an+1

整理,得class="stub"2
an+1
=class="stub"1
an
+class="stub"1
an+2

{class="stub"1
an
}
是等差数列.
∵a1=class="stub"1
4
,a2=class="stub"1
5

∴等差数列{class="stub"1
an
}
的首项是class="stub"1
a1
=4
,公差d=class="stub"1
a2
-class="stub"1
a1
=5-4=1

class="stub"1
an
=4+(n-1)×1=n+3

class="stub"1
a1
+class="stub"1
a2
+…+class="stub"1
a97
=97× 4+class="stub"97×96
2
×1
=5044.
故选B.

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