已知递增的等差数列{an}满足:a2a3=45,a1+a4=14(1)求数列{an}的通项公式及前n项和Sn;(2)设bn=an+1Sn,求数列{bnbn+1}的前n项和Tn.-数学

题目简介

已知递增的等差数列{an}满足:a2a3=45,a1+a4=14(1)求数列{an}的通项公式及前n项和Sn;(2)设bn=an+1Sn,求数列{bnbn+1}的前n项和Tn.-数学

题目详情

已知递增的等差数列{an}满足:a2a3=45,a1+a4=14
(1)求数列{an}的通项公式及前n项和Sn
(2)设bn=
an+1
Sn
,求数列{bnbn+1}的前n项和Tn
题型:解答题难度:中档来源:不详

答案

(1)由题意得,a1+a4=14,则a2+a3=14,
∵a2a3=45,∴a2、a3是方程x2-14x+45=0的两根,
∵等差数列{an}是递增数列,∴a2<a3,
解得a2=5,a3=9,公差d=4,a1=1,
∴an=4n-3,
Sn=
n(a1+an)
2
=
n(1+4n-3)
2
=2n2-n,
(2)由(1)得,bn=
an+1
Sn
=class="stub"4n-2
2n2-n
=class="stub"2
n

则bn•bn+1=class="stub"4
n(n+1)
=4(class="stub"1
n
-class="stub"1
n+1
),
∴Tn=b1•b2+b2•b3+…+bn•bn+1
=4[(1-class="stub"1
2
)+(class="stub"1
2
-class="stub"1
3
)+…+(class="stub"1
n
-class="stub"1
n+1
)]
=4(1-class="stub"1
n+1
)=class="stub"4n
n+1

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