已知正项数列an的前n项和为sn,且满足：an平方=2sn-an（n属于N*).求an的通项公式；2.求数列｛an,2an(此an
问题描述：
已知正项数列an的前n项和为sn,且满足：an平方=2sn-an（n属于N*).求an的通项公式；2.求数列｛an,2an(此an
答
1 an^2=2sn-an
an-1^2=2sn-1-an-1
an^2-an-1^2=2(sn-sn-1)-an+an-1
an^2-an-1^2=2an-an+an-1
(an+an-1)(an-an-1)=an+an-1
an-an-1=1(易知an+an-1不等于0)
a1^2=2a1-a1 a1=1
an=1+1*(n-1)=n
2 题目没打完吧
答
(An)^2=2Sn-An=>(A(n-1))^2=2S(n-1)-A(n-1)=>(An)^2-(A(n-1))^2=2Sn-An-2S(n-1)+A(n-1)=>(An+A(n-1))*(An-A(n-1))=2An-An+A(n-1)=>(An+A(n-1))*(An-A(n-1))=An+A(n-1)正项数列=>An+A(n-1)=0不成立=>An-A(n-1)=1又A1=...