已知数列{an}的前n项和为Sn,且满足a1=1,2an/(anSn-Sn^2)=1(n大于等于2)

问题描述:

已知数列{an}的前n项和为Sn,且满足a1=1,2an/(anSn-Sn^2)=1(n大于等于2)
(1)证明数列{1/Sn}成等差数列,并求Sn
(2)求数列{an}的通项公式

由题意知:2an/[anSn-(Sn)²]=1(n>1)则:(Sn)²-anSn+2an=0(n>1)又因为:an=Sn-S(n-1)(n>1)所以:(Sn)²-[Sn-S(n-1)]Sn+2[Sn-S(n-1)]=0展开化简:S(n-1)Sn+2Sn-2S(n-1)=0两边同除以S(n-1)Sn,得:1...