正项数列{an}满足a2n-(2n-1)an-2n=0.(1)求数列{an}的通项公式an;(2)令bn=1(n+1)an,求数列{bn}的前n项和Tn.-数学

题目简介

正项数列{an}满足a2n-(2n-1)an-2n=0.(1)求数列{an}的通项公式an;(2)令bn=1(n+1)an,求数列{bn}的前n项和Tn.-数学

题目详情

正项数列{an}满足
a2n
-(2n-1)an-2n=0.
(1)求数列{an}的通项公式an
(2)令bn=
1
(n+1)an
,求数列{bn}的前n项和Tn
题型:解答题难度:中档来源:江西

答案

(1)由正项数列{an}满足:
a2n
-(2n-1)an-2n=0,
可得(an-2n)(an+1)=0
所以an=2n.
(2)因为an=2n,bn=class="stub"1
(n+1)an

所以bn=class="stub"1
(n+1)an

=class="stub"1
2n(n+1)

=class="stub"1
2
(class="stub"1
n
-class="stub"1
n+1
)

Tn=class="stub"1
2
(1-class="stub"1
2
+class="stub"1
2
-class="stub"1
3
+…+class="stub"1
n
-class="stub"1
n+1
)

=class="stub"1
2
(1-class="stub"1
n+1
)

=class="stub"n
2n+2

数列{bn}的前n项和Tn为class="stub"n
2n+2

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