x(n+1) +1=2xn^2+4xn+2=2(xn+1)^2
两边取对数得
lg[x(n+1) +1]=lg2+2lg(xn+1)
lg[x(n+1) +1]+lg2=2[lg(xn+1)+lg2]
{lg(xn+1)+lg2}是等比数列,lg(x1+1)+lg2=2lg2
lg(xn+1)+lg2=2lg2*2^(n-1)=2^nlg2
2(xn+1)=2^(2^n)
xn=2^(2^n-1)-1