正整数{an}的前n项和为Sn,且an=2√ Sn-1,求数列{an}的通项公式

a1=2√ S1-1a1=2√a1-1a1-2√a1+1=0(√a1-1)^2=0a1=1an=2√ Sn-12√ Sn=an+14Sn=(an+1)^24S(n-1)=[a(n-1)+1]^2以上两式相减得4Sn-4S(n-1)=(an+1)^2-[a(n-1)+1]^24an=(an+1)^2-[a(n-1)+1]^24an=(an)^2+2an+1-[a(n-1)]...