一直点（an,an+1）在函数y=x²+2x的图像上,其中n=1,2,3.证明数列lg（1+an）是等比数列

a(n+1) = [a(n)]^2 + 2a(n),a(n+1) + 1 = [a(n)]^2 + 2a(n) + 1 = [a(n)+1]^2,若a(n+1)+1 = 0,则a(n) + 1 = 0 ,...,a(1) + 1 = 0,a(1) = -1,a(n) = -1.若a(1)不为－1,则,a(n) + 1不为0.因此,a(n+1) + 1 = [a(n) + 1]...