2根号Ｓn =an+1
4Sn=(an+1)^
4S(n+1) =[a(n+1)+1]^
4a(n+1) =[a(n+1)]^-(an)^+2a(n+1)-2an
====>[a(n+1)]^-(an)^-2a(n+1)-2an=0
[a(n+1)+an][a(n+1)-an -2]=0
因为是正数数列 a(n+1)+an >0
===> 只有 a(n+1)-an -2 =0
a(n+1) =an+2
又
2根号S1 =a1+1 ===>a1=1
===>an =1+(n-1)*2 =2n -1
答：解: a1=S1=2a1-3,得a1=3 Sn=2an-3n S(n-1)=2a(n-1)-3(n-1) Sn-S(n-1)=2an-2a(n-1)-3=an ...详情>>
答：详情>>
