定义:若数列{An}满足An+1=An2,则称数列{An}为“平方递推数列”.

1.
x=an f(x)=a(n+1)代入f(x)=2x²+2x
a(n+1)=2an²+2an
2a(n+1)+1=4an²+4an+1=(2an +1)²
2a1+1=2×2+1=5
满足平方递推数列定义,因此数列{2an +1}是以5为首项的平方递推数列.
2an +1=[2a(n-1)+1]²=[2a(n-2)+1]⁴=...=(2a1+1)^[2^(n-1)]=5^[2^(n-1)]
lg(2an +1)=lg[5^(2n-2)]=[2^(n-1)]lg5
lg[2a(n+1)+1]/[lg(2an +1)]=2^(n+1-1)lg5/[2^(n-1) lg5]=2,为定值.
lg(2a1+1)=lg5
数列{lg(2an +1)}是以lg5为首项,2为公比的等比数列.
2.
由1得2an +1=5^[2^(n-1)]
an={5^[2^(n-1)] -1}/2
Tn=(2a1+1)(2a2+1)...(2an+1)
=5^1×5^2×5^(2²)×...×5^[2^(n-1)]
=5^[1+2+2²+...+2^(n-1)]
=5^[1×(2ⁿ-1)/(2-1)]
=5^(2ⁿ-1)
3.
bn=log(1+2an)(Tn)
=log[5^(2^(n-1))][5^(2ⁿ-1)]
=lg[5^(2ⁿ-1)]/lg[5^[2^(n-1)]]
=(2ⁿ-1)lg5/[2^(n-1) lg5]
=(2ⁿ-1)/2^(n-1)
=2 -1/2^(n-1)
Sn=b1+b2+...+bn
=2n -[1/2^0+1/2+...+1/2^(n-1)]
=2n-1×(1-1/2ⁿ)/(1-1/2)
=2n-2+2/2ⁿ
Sn>2012
2n-2+2/2ⁿ>2012
n+1/2ⁿ>1007
n>1007 -1/2ⁿ
01007
n的最小值为1007.