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定义:在数列中,若,(n≥2,n∈N*,p为常数),则称为“等方差数列”.下列是...

定义:在数列6ec8aac122bd4f6e中,若6ec8aac122bd4f6e,(n≥2,n∈N*,p为常数),则称6ec8aac122bd4f6e为“等方差数列”.下列是对“等方差数列”的有关判断:

①若6ec8aac122bd4f6e是“等方差数列”,则数列6ec8aac122bd4f6e是等差数列;②6ec8aac122bd4f6e是“等方差数列”;

③若6ec8aac122bd4f6e是“等方差数列”,则数列6ec8aac122bd4f6e(k∈N*,k为常数)也是“等方差数列”;

④若6ec8aac122bd4f6e既是“等方差数列”,又是等差数列,则该数列是常数数列.

其中正确的命题为                 .(写出所有正确命题的序号)

 

③④ 【解析】因为①:可以举反例.如an=0时数列不存在,所以①错误 ②:对数列{(-2)n}有an2-an-12=[(-2)n]2-[(-2)n-1]2=4n-4n-1不是常数,所以②错误 ③:对数列{akn}有akn2-ak(n-1)2=(akn2-akn-12)+(akn-12-akn-22)+…+(akn-k+12-akn-k2)=kp,而k,p均为常数,所以数列{akn}也是“等方差数列”,所以③正确 ④:设数列{an}首项a1,公差为d则有a2=a1+d,a3=a1+2d,所以有(a1+d)2-a12=p,且(a1+2d)2-(a1+d)2=p,所以得d2+2a1d=p,3d2+2a1d=p,上两式相减得d=0,所以此数列为常数数列,所以④正确. 故答案为:③④
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