已知函数f(x)=-4+1x2,数列{an},点Pn(an,-1an+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

题目简介

已知函数f(x)=-4+1x2,数列{an},点Pn(an,-1an+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

题目详情

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
题型:解答题难度:中档来源:不详

答案

(1)-class="stub"1
an+1
=f(an)=-
4+class="stub"1
a2n
且an>0
class="stub"1
an+1
=
4+class="stub"1
a2n
class="stub"1
a2n+1
-class="stub"1
a2n
=4

∴数列{class="stub"1
a2n
}是等差数列,首项class="stub"1
a21
=1
,公差d=4
class="stub"1
a2n
=1+4(n-1)
a2n
=class="stub"1
4n-3

∵an>0∴an=
class="stub"1
4n-3

(2)bn=class="stub"1
4n-3
•class="stub"1
4n+1
=class="stub"1
4
(class="stub"1
4n-3
-class="stub"1
4n+1

Tn=class="stub"1
4
(1-class="stub"1
5
+class="stub"1
5
-class="stub"1
9
+…+class="stub"1
4n-3
-class="stub"1
4n+1
)
=class="stub"1
4
(1-class="stub"1
4n+1
)
=class="stub"n
4n+1

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