(1) 证明： 因为 Sn=k*an,故 S(n-1) = k*a(n-1)
所以 Sn－S(n-1)＝k*an - k*[a(n-1) = k*[an-a(n-1)
即有 an = k*[an - a(n-1),整理得 an/a(n-1) = k/(k-1)
因此,当k≠1,0时,数列{ an }是等比数列
(2) 由(1)可知
an = a1*[k/(k-1)]^(n-1)
(3) 当k=-1时,数列{an}的公比q = k/(k-1)=1/2
a2^2 = (a1q)^2 = (1/4)a1^2,a3^2 = (a1q^2)^2 = (1/4)^2,(an) = (a1q^(n-1))^2 = a1^2*(1/4)^(n-1)
所以 a1^2 + a2^2 + a3^2 +……+an^2 = a1^2*[1 + (1/4) +(1/4)^2 +……+(1/4)^(n-1)]
= a1^2*{1*[1-(1/4)^n]/[1-(1/4)]
= [(4^n - 1)*a1^2]/[3*4^(n-1)]