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,2a

(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=1
=>
An=n
Bn=An*2^(An)
=>
Bn=n*2^n
=>
Sn=1*2^1+2*2^2+3*2^3+...n*2^n
=>
2Sn=0+1*2^2+2*2^3+3*2^4+...+n*2^(n+1)
=>
Sn=0+1*2^2+2*2^3+3*2^4+...+n*2^(n+1)-(1*2^1+2*2^2+3*2^3+...n*2^n)
=>
Sn
=n*2^(n+1)-1*2^1-1*2^2-1*2^3-...-1*2^n
=n*2^(n+1)-2(1-2^n)/(1-2)
=n*2^(n+1)+2(1-2^n)
Bn=3^n+(-1)^(n-1)*x*2^An
=>
Bn=3^n+(-1)^(n-1)*x*2^n
=>
B(n+1)=3^(n+1)+(-1)^n*x*2^(n+1)
B(n+1)>Bn
=>
3^(n+1)+(-1)^n*x*2^(n+1)>3^n+(-1)^(n-1)*x*2^n
=>
3^(n+1)-3^n>(-1)^(n-1)*x*2^n-(-1)^n*x*2^(n+1)
=>
2*3^n>(-1)^(n-1)*x*2^n+(-1)^(n+1)*x*2^(n+1)
=>
2*3^n>(-1)^(n-1)*x*2^n*3
恒成立
=>
2*3^n>x*2^n*3
=>
3^(n-1)>x*2^(n-1)
=>
(1.5)^(n-1)>x恒成立
=>
x

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