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

(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=...