若数列{An}满足An+1=An^2,则称数列{An}为“平方递推数列”,已知数列{an}中,a1=9,点（an,an+1）在函数f（x）=x^2+2x的图像上,其中n为正整数,

x=an f(x)=a(n+1)代入函数方程
a(n+1)=an^2 +2an
a(n+1) +1=an^2+2an +1=(an +1)^2
满足平方递推数列定义,因此数列{an +1}是平方递推数列.
a1+1=10>0,若当n=k(k∈N+)时,ak>0,则a(k+1)=ak^2>0,k为任意正整数,因此对于任意正整数n,数列{an +1}各项恒为正.
lg(a1+1)=lg(9+1)=lg10=1
an +1=[a(n-1)+1]^2
=[a(n-2)+1]^(2^2)
=[a(n-3)+1]^(2^3)
=.
=[a1+1]^[2^(n-1)]
=(9+1)^[2^(n-1)]
=10^[2^(n-1)]
lg(an +1)=lg[10^[2^(n-1)]]=2^(n-1)
lg[a(n+1)+1]/(an +1)=2^n/2^(n-1)=2,为定值,数列{lg(an +1)}是以1为首项,2为公比的等比数列
lg(an +1)=1×2^(n-1)=2^(n-1)
lg(Tn)=lg[(a1+1)(a2+1)...(an +1)]
=lg(a1+1)+lg(a2+1)+...+lg(an +1)
=1×(2ⁿ -1)/(2-1) /这一步不难理解吧,就是等比数列求和
=2ⁿ-1