已知数列{an},{bn}是各项均为正数的等比数列,设cn=bnan,n∈N*(Ⅰ)证明:数列{cn}是等比数列,数列{lnan}是等差数列.(Ⅱ)设数列{lnan},{lnbn}的前n项和分别是Sn

题目简介

已知数列{an},{bn}是各项均为正数的等比数列,设cn=bnan,n∈N*(Ⅰ)证明:数列{cn}是等比数列,数列{lnan}是等差数列.(Ⅱ)设数列{lnan},{lnbn}的前n项和分别是Sn

题目详情

已知数列{an},{bn}是各项均为正数的等比数列,设cn=
bn
an
,n∈N*
(Ⅰ)证明:数列{cn}是等比数列,数列{lnan}是等差数列.
(Ⅱ)设数列{lnan},{lnbn}的前n项和分别是Sn,Tn.若a1=2,
Sn
Tn
=
n
2n+1
,求数列{cn}的通项公式.
(Ⅲ)在(Ⅱ)条件下,设dn=
6cn
bn+1-4an+1-4an+2  
,求数列{dn}的前n项和.
题型:解答题难度:中档来源:不详

答案

(1)设数列{an}、bn的公比分别为p、q(p>0,q>0),
则由题意可得
an
an-1
=p,
bn
bn-1
=q

cn
cn-1
=
anbn
an-1bn-1
= pq
,c1=a1•b1
所以数列cn以a1•b1为首项,以pq为公比的等比数列
又因为lnan-lnan-1=ln
an
an-1
=lnp

数列lnan以lna1为首项,以lnp为公差的等差数列
(2)由题意可得sn=n•ln2+
n(n-1)
2
×lnp
Tn=n•lnb1+
n(n-1)
2
×lnq

Sn
Tn
=
n•ln2+
n(n-1)
2
• lnp
n•lnb1+
n(n-1)
2
•lnq
=
2ln2+(n-1)•lnp
2lnb1+(n-1)•lnq
=class="stub"n
2n+1

n•lnp+(ln4-lnp)
n•lnq+(2lnb1-lnq)
=
n+class="stub"ln4-lnp
lnp
n•class="stub"lnq
lnp
+
2lnb1-lnq
lnp
=class="stub"n
2n+1

class="stub"ln4-lnp
lnp
=0,class="stub"lnq
lnp
=2,
2lnb1-lnq
lnp
=1

∴p=4,q=16,b1=8
∴an=2•4n-1=22n-1,bn=8•16n-1=24n-1
(III)由(II)可得dn=
6•cn
bn+1-4an+1-4an+2

=
6•4n
8•16n-8•4n-2•4n+2

=
3•4n
4•(4n)2-5•4n+1

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

∴d1+d2+d3+…+dn
=class="stub"1
41-1
-class="stub"1
42-1
+class="stub"1
42-1
-class="stub"1
43-1
+…+class="stub"1
4n-1
-class="stub"1
4n+1-1

=class="stub"1
3
-class="stub"1
4n+1-1

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