数列{an}满足a1=2,an+1=an^2+6an+6

a(n+1)+3=an²+6an+9=(an+3)²
取对数
lg[a(n+1)+3]=2lg(an+3)
所以lg(an+3)是等比数列，q=2
则lg(an+3)=lg(a1+3)*2^(n-1)=2^(n-1)*lg5
即lg(an+3)=lg5^2^(n-1)
所以an+3=5^[2^(n-1)]
an=-3+5^[2^(n-1)]

a(n+1)=an^2+6an+6
a(n+1)+3=an^2+6an+9
a(n+1)+3=(an+3)^2
取对数
lg[a(n+1)+3]=lg[(an+3)^2]
lg[a(n+1)+3]=2lg[(an+3)]
lg[a(n+1)+3]/lg[(an+3)]=2
所以lg[(an+3)]是以2为公比的等比数列
lg[(an+3)]=lg[(a1+3)]*q^(n-1)
an+3=(a1+3)^[q^(n-1)]
an+3=(2+3)^[q(n-1)]
an+3=5^[2(n-1)]
an+3=5^(2n-2)
an=5^(2n-2)-3

a(n+1)=(an)^2+6an+6 两边同加3得到
a(n+1)+3=(an)^2+6an+9=(an+3)^2;
所以an+3=[a(n-1)+3]^2=[a(n-2)+3]^4=……=(a1+3)^(2^(n-1))=5^(2^(n-1));

所以 an=5^(2^(n-1))-3;

an+1=an^2+6an+6=(an+3)^2-3
a(n+1)+3=(an+3)^2
lg[a(n+1)+3]=lg[(an+3)^2]
lg[a(n+1)+3]/lg(an+3)=2 等比 首项 lg(a1+3)=lg5
lg(an+3)=lg5*2^(n-1)
an+3=10^lg5*2^(n-1)
an=10^lg5*2^(n-1)-3
ok

an+1 + 3=an^2+6an+9=(an+3)^2
所以a2+3=(a1+3)^2=5^2,所以a2=5^2-3
a3+3=(a2+3)^2=5^4,所以a3=5^4-3
跌代可得
an=5^2^(n-1)-3