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.
问题描述:
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{
}是等差数列,并求公差;1 Sn
(2)求数列{an}的通项公式.
答
(1)∵2an=SnSn-1(n≥2)∴2(Sn-Sn-1)=SnSn-1两边同时除以SnSn-1,得2(1Sn-1-1Sn)=1∴1Sn-1Sn-1=-12∴{1Sn}是等差数列,公差d=-12(2)∵1S1=1a1=13∴1Sn=13+(n-1)×(-12)=-12n+56=5-3n6∴Sn=65-3n当n≥2时,an=...