Question

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

Answer 1

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]^2 > 0,

a(n) + 1 > 0.

lg[1 + a(n+1)] = lg[1+ a(n)]^2 = 2lg[1 + a(n)],

{lg[1+a(n)]}是首项为lg[1+a(1)],公比为2的等比数列.