已知数列{an}是首项为a1=14,公比q=14的等比数列.设bn+2=3log14an(n∈N*),数列{cn}满足cn=1bn•bn+1.(Ⅰ)求证:数列{bn}成等差数列;(Ⅱ)求数列{cn}的

题目简介

已知数列{an}是首项为a1=14,公比q=14的等比数列.设bn+2=3log14an(n∈N*),数列{cn}满足cn=1bn•bn+1.(Ⅰ)求证:数列{bn}成等差数列;(Ⅱ)求数列{cn}的

题目详情

已知数列{an}是首项为a1=
1
4
,公比q=
1
4
的等比数列.设bn+2=3log
1
4
an
(n∈N*),数列{cn}满足cn=
1
bnbn+1

(Ⅰ)求证:数列{bn}成等差数列;
(Ⅱ)求数列{cn}的前n项和Sn
题型:解答题难度:中档来源:不详

答案

证明:(Ⅰ)∵数列{an}是首项为a1=class="stub"1
4
,公比q=class="stub"1
4
的等比数列,
∴an=class="stub"1
4
(class="stub"1
4
)
n-1
=(class="stub"1
4
)
n

∵bn+2=3logclass="stub"1
4
an=3logclass="stub"1
4
(class="stub"1
4
)
n
=3n(n∈N*),
∴bn=3n-2;
∴bn+1-bn=3(n+1)-2-(3n-2)=3,
∴数列{bn}是以1为首项,3为公差的成等差数列.
(Ⅱ)∵cn=class="stub"1
bn•bn+1
=class="stub"1
(3n-2)[3(n+1)-2]
=class="stub"1
3
class="stub"1
3n-2
-class="stub"1
3n+1
),
∵数列{cn}的前n项和为Sn,
∴Sn=c1+c2+…+cn
=class="stub"1
3
[(1-class="stub"1
4
)+(class="stub"1
4
-class="stub"1
7
)+…+(class="stub"1
3n-2
-class="stub"1
3n+1
)]
=class="stub"1
3
(1-class="stub"1
3n+1

=class="stub"n
3n+1

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