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=an•bn.(1)求证:{bn}是等差数列;(2)求数列{cn}的前n项和S

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=an•bn.(1)求证:{bn}是等差数列;(2)求数列{cn}的前n项和S

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=an•bn
(1)求证:{bn}是等差数列;
(2)求数列{cn}的前n项和Sn
(3)若Cn
1
4
m2+m-1
对一切正整数n恒成立,求实数m的取值范围.
题型:解答题难度:中档来源:不详

答案

(1)由题意知,an=(class="stub"1
4
)n.
bn+2=3logclass="stub"1
4
an
b1+2=3logclass="stub"1
4
a1

∴b1=1
∴bn+1-bn=3logclass="stub"1
4
an+1=3logclass="stub"1
4
an=3logclass="stub"1
4
an+1
a n
=3logclass="stub"1
4
q=3
∴数列{bn}是首项为1,公差为3的等差数列.
(2)由(1)知,an=(class="stub"1
4
)n.bn=3n-2
∴Cn=(3n-2)×(class="stub"1
4
)n.
∴Sn=1×class="stub"1
4
+4×(class="stub"1
4
)2+…+(3n-2)×(class="stub"1
4
)n,
于是class="stub"1
4
Sn=1×(class="stub"1
4
)2+4×(class="stub"1
4
)3+…(3n-2)×(class="stub"1
4
)n+1,
两式相减得class="stub"3
4
Sn=class="stub"1
4
+3×[(class="stub"1
4
)2+(class="stub"1
4
)3+…+(class="stub"1
4
)n)-(3n-2)×(class="stub"1
4
)n+1,
=class="stub"1
2
-(3n-2)×(class="stub"1
4
)n+1,
∴Sn=class="stub"2
3
-class="stub"12n+8
3
×
class="stub"1
4
)n+1
(3)∵Cn+1-Cn=(3n+1)×(class="stub"1
4
)n+1-(3n-2)×(class="stub"1
4
)n=9(1-n)×(class="stub"1
4
)n+1,
∴当n=1时,C2=C1=class="stub"1
4

当n≥2时,Cn+1<Cn,即C2=C1>C3>C4<…>Cn
∴当n=1时,Cn取最大值是class="stub"1
4

Cn≤class="stub"1
4
m2+m-1

class="stub"1
4
m2+m-1
class="stub"1
4

即m2+4m-5≥0解得m≥1或m≤-5.

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