已知数列{an}的前n项和Sn=12n(n-1),且an是bn与1的等差中项.(1)求数列{an}和数列{bn}的通项公式;(2)令cn=an3n,求数列{Cn}的前n项和Tn;(3)若f(n)=an

题目简介

已知数列{an}的前n项和Sn=12n(n-1),且an是bn与1的等差中项.(1)求数列{an}和数列{bn}的通项公式;(2)令cn=an3n,求数列{Cn}的前n项和Tn;(3)若f(n)=an

题目详情

已知数列{an}的前n项和Sn=
1
2
n(n-1)
,且an是bn与1的等差中项.
(1)求数列{an}和数列{bn}的通项公式;
(2)令cn=
an
3n
,求数列{Cn}的前n项和Tn
(3)若f(n)=
an(n=2k-1)
bn(n=2k)
(k∈N*),是否存在n∈N*,使得f(n+13)=2f(n),并说明理由.
题型:解答题难度:中档来源:不详

答案

(1)由Sn=class="stub"1
2
n2-class="stub"1
2
n
,由an=
S1n=1
Sn-Sn-1n≥2

求得an=n-1
又∵2an=bn+1
∴bn=2n-3
(2)Cn=class="stub"n-1
3n

Tn=0×(class="stub"1
3
)+1•(class="stub"1
3
)2++(n-1)•(class="stub"1
3
)n
class="stub"1
3
Tn=0•(class="stub"1
3
)2++(n-2)(class="stub"1
3
)n+(n-1)•(class="stub"1
3
)n+1

两式相减得:class="stub"2
3
Tn=1×(class="stub"1
3
)2++(class="stub"1
3
)n-(n-1)•(class="stub"1
3
)n+1

class="stub"2
3
Tn=
(class="stub"1
3
)
2
•[1-(class="stub"1
3
)
n-1
]
1-class="stub"1
3
-(n-1)•(class="stub"1
3
)n+1=class="stub"1
6
•[1-class="stub"1
3n-1
]-class="stub"n-1
3n+1

Tn=class="stub"1
4
-class="stub"1
4
•class="stub"1
3n-1
-class="stub"n-1
2•3n
=class="stub"1
4
-class="stub"2n+1
4•3n

(3)当n为奇数时:f(n)=an=n-1f(n+13)=2n+23
∴2n+23=2n-2⇒n∈ϕ
当n为偶数时f(n)=bn=2n-3f(n+13)=n+12由题
∴2•(2n-3)=n+12⇒n=6为偶数
∴满足条件的n存在且等于6.

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