问题 解答题
已知函数f(x)=-
4+
1
x2
,数列{an},点Pn(an,-
1
an+1
)在曲线y=f(x)上(n∈N+),且a1=1,an>0.
( I)求数列{an}的通项公式;
( II)数列{bn}的前n项和为Tn且满足bn=an2an+12,求Tn
答案

(1)-

1
an+1
=f(an)=-
4+
1
a2n
且an>0

1
an+1
=
4+
1
a2n
1
a2n+1
-
1
a2n
=4

∴数列{

1
a2n
}是等差数列,首项
1
a21
=1
,公差d=4

1
a2n
=1+4(n-1)∴
a2n
=
1
4n-3

∵an>0∴an=

1
4n-3

(2)bn=

1
4n-3
1
4n+1
=
1
4
(
1
4n-3
-
1
4n+1

Tn=

1
4
(1-
1
5
+
1
5
-
1
9
+…+
1
4n-3
-
1
4n+1
)=
1
4
(1-
1
4n+1
)
=
n
4n+1

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