设数列{an}的前n项和Sn满足:Sn=nan-2n(n-1).等比数列{bn}的前n项和为Tn,公比为a1,且T5=T3+2b5.(1)求数列{an}的通项公式;(2)设数列{1anan+1}的前n

题目简介

设数列{an}的前n项和Sn满足:Sn=nan-2n(n-1).等比数列{bn}的前n项和为Tn,公比为a1,且T5=T3+2b5.(1)求数列{an}的通项公式;(2)设数列{1anan+1}的前n

题目详情

设数列{an}的前n项和Sn满足:Sn=nan-2n(n-1).等比数列{bn}的前n项和为Tn,公比为a1,且T5=T3+2b5
(1)求数列{an}的通项公式;
(2)设数列{
1
anan+1
}的前n项和为Mn,求证:
1
5
≤Mn
1
4
题型:解答题难度:中档来源:不详

答案

(1)∵等比数列{bn}的前n项和为Tn,公比为a1,且T5=T3+2b5 ,∴b4+b5=2b5,
∴b4=b5,∴公比 a1=
b5
b4
=1,故等比数列{bn}是常数数列.
数列{an}的前n项和Sn满足:Sn=nan-2n(n-1),当n≥2时,
an=sn-sn-1=nan-2n(n-1)-[nan-1-2(n-1)(n-2)],∴an-an-1=4 (n≥2).
∴数列{an}是以1为首项,以4为公差的等差数列,an=4n-3.
(2)∵数列{class="stub"1
anan+1
}的前n项和为Mn,
class="stub"1
anan+1
=class="stub"1
(4n-3)[4(n+1)-3]
=class="stub"1
(4n-3)(4n+1)
=class="stub"1
4
(class="stub"1
4n-3
-class="stub"1
4n+1
)

∴Mn =class="stub"1
4
[1-class="stub"1
5
+class="stub"1
5
-class="stub"1
9
+class="stub"1
9
-class="stub"1
13
+…+class="stub"1
4n-3
-class="stub"1
4n+1
]=class="stub"1
4
(1-class="stub"1
4n+1
)<class="stub"1
4

再由数列{ Mn }是增数列,∴Mn≥M1=class="stub"1
5

综上可得,class="stub"1
5
≤Mn<class="stub"1
4

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