已知数列{an}的首项a1=3,通项an与前n项和Sn之间满足2an=SnSn-1(n≥2).(1)求证{1Sn}是等差数列,并求公差;(2)求数列{an}的通项公式.-数学

题目简介

已知数列{an}的首项a1=3,通项an与前n项和Sn之间满足2an=SnSn-1(n≥2).(1)求证{1Sn}是等差数列,并求公差;(2)求数列{an}的通项公式.-数学

题目详情

已知数列{an}的首项a1=3,通项an与前n项和Sn之间满足2an=SnSn-1(n≥2).
(1)求证{
1
Sn
}
是等差数列,并求公差;
(2)求数列{an}的通项公式.
题型:解答题难度:中档来源:不详

答案

(1)∵2an=SnSn-1(n≥2)∴2(Sn-Sn-1)=SnSn-1
两边同时除以SnSn-1,得2(class="stub"1
Sn-1
-class="stub"1
Sn
)=1

class="stub"1
Sn
-class="stub"1
Sn-1
=-class="stub"1
2

{class="stub"1
Sn
}
是等差数列,公差d=-class="stub"1
2

(2)∵class="stub"1
S1
=class="stub"1
a1
=class="stub"1
3

class="stub"1
Sn
=class="stub"1
3
+(n-1)×(-class="stub"1
2
)=-class="stub"1
2
n+class="stub"5
6
=class="stub"5-3n
6

Sn=class="stub"6
5-3n

当n≥2时,an=class="stub"1
2
SnSn-1=class="stub"1
2
×class="stub"6
5-3n
×class="stub"6
8-3n
=class="stub"18
(5-3n)(8-3n)

an=
3,n=1
class="stub"18
(8-3n)(5-3n)
,n≥2

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