设数列{an}的首项a1=32,前n项和为Sn,且满足2an+1+Sn=3(n∈N*).(Ⅰ)求a2及an;(Ⅱ)求满足1817<S2nSn<87的所有n的值.-数学

题目简介

设数列{an}的首项a1=32,前n项和为Sn,且满足2an+1+Sn=3(n∈N*).(Ⅰ)求a2及an;(Ⅱ)求满足1817<S2nSn<87的所有n的值.-数学

题目详情

设数列{an}的首项a1=
3
2
,前n项和为Sn,且满足2an+1+Sn=3( n∈N*).
(Ⅰ)求a2及an
(Ⅱ)求满足
18
17
S2n
Sn
8
7
的所有n的值.
题型:解答题难度:中档来源:不详

答案

(Ⅰ)由2an+1+Sn=3,得2a2+a1=3,
a1=class="stub"3
2
,所以a2=class="stub"3
4

由2an+1+Sn=3,2an+Sn-1=3(n≥2)相减,
an+1
an
=class="stub"1
2

a2
a1
=class="stub"1
2
,所以数列{an}是以class="stub"3
2
为首项,
class="stub"1
2
为公比的等比数列.
因此an=class="stub"3
2
•(class="stub"1
2
)n-1=3•(class="stub"1
2
)n
(n∈N*).
(Ⅱ)由题意与(Ⅰ),
class="stub"18
17
S2n
Sn
=1+(class="stub"1
2
)n<class="stub"8
7

class="stub"1
17
<(class="stub"1
2
)n<class="stub"1
7

因为class="stub"1
17
<(class="stub"1
2
)3<class="stub"1
7
class="stub"1
17
<(class="stub"1
2
)4<class="stub"1
7

所以n的值为3,4.

更多内容推荐