若{an}是等方差数列,则满足(an)^2-(a(n-1))^2 = p(n>=2,p为常数)
令bn = (an)^2,则得到:bn- b(n-1) = p (n>=2,p为常数) 所以{bn}为等差数列,也即
{(an)^2}为等差数列。
设an = (-1)^n
则(an)^2-(a(n-1))^2 = [(-1)^n]^2-[(-1)^(n-1)]^2 = [(-1)^2]^n-[(-1)^2]^(n-1) =1^n - 1^(n-1) =1-1=0
所以{ (-1)^n}为等方差数列。
若{an}是等方差数列,则满足(a(kn))^2-(a(kn-1))^2 = p(n>=2,p为常数)
令bn = a(kn),则得到:(bn)^2- (b(n-1))^2 = (a(kn))^2- (a(k(n-1))^2 = (a(kn))^2- (a(kn-k))^2
=(a(kn))^2-(a(k(n-1))^2+(a(k(n-1))^2-(a(k(n-2))^2+(a(k(n-2))^2-...-(a(k(n-(k-1)))^2+(a(k(n-(k-1)))^2+ (a(kn-k))^2
=p+p +...+p(一共k个) = kp 为一常量,所以{bn}满足等方差数列的定义,为等方差数列,也即{a(kn)}为等方差数列。