问题 选择题
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
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

n+1
an+1
-
n
an+2
=4,

同理,得

n
an
-
n-1
an+1
=4,

n+1
an+1
-
n
an+2
=
n
an
-
n-1
an+1

整理,得

2
an+1
=
1
an
+
1
an+2

{

1
an
}是等差数列.

∵a1=

1
4
,a2=
1
5

∴等差数列{

1
an
}的首项是
1
a1
=4
,公差d=
1
a2
-
1
a1
=5-4=1

1
an
=4+(n-1)×1=n+3.

1
a1
+
1
a2
+…+
1
a97
=97× 4+
97×96
2
×1
=5044.

故选B.

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