数列{an}中,a1=1,当n>1时,2Sn^2=2anSn-an,求通项an
问题描述：
数列{an}中,a1=1,当n>1时,2Sn^2=2anSn-an,求通项an
答
2Sn^2=2anSn-an=an(2Sn-1)=(Sn-S(n-1))(2Sn-1)=2Sn^2-2SnS(n-1)-Sn+S(n-1)
2SnS(n-1)=-Sn+S(n-1)
1/Sn-1/S(n-1)=2
{1/Sn}是以2为公差的等差数列
则1/Sn=S1+2（n-1）=2n-1
很容易an=-2/(2n-1)(2n-3)
答
当n>1时,an=sn-sn-1代入化简得：
1/Sn-1/Sn-1=2
所以：1/Sn=2n-1
所以：Sn=1/(2n-1)
当n>1时,an=sn-sn-1
=-2/[(2n-1)(2n-3)]
当n=1时a1=1也符合
an=-2/[(2n-1)(2n-3)]