设数列{an}中,a1=1,且n大于1时,2Sn^2=2anSn—an求an

题目

设数列{an}中,a1=1,且n大于1时,2Sn^2=2anSn—an求an
看了答案之后卡在了1/(Sn+1)-1/Sn=2,1/S(1)=1/A(1)=1/2为首项,为什么代n=1进去呢?,2Sn^2=2anSn—an规定了n大于1啊?

答案

an=sn-s(n-1),2sn^2=2(sn-sn-1)sn-sn+s(n-1)=2sn^2-2s(n-1)sn-sn+s(n-1) 2sns(n-1)=s(n-1)-sn
2=1/sn-1/s(n-1) n=2,2=1/s2-1/s1,n=3,2=1/s3-1/s2,n=4,2=1/s4-1/s3,.2=1/sn-1/s(n-1),相加即得1/sn-1/s1=2(n-1),又a1=s1=1,所以1/sn=2(n-1)+1/2=2n-3/2,sn=1/(2n-3/2)=2/(4n-3).(n>1)时an=sn-s(n-1)=2/(4n-3)-2/(4n-7) ．当n=1时,a1=2