Question

已知{an}为等差数列,公差为d,求证{C^an}(c不等于0)为等比数列并写出公比（要过程）

Answer 1

{an}为等差数列,公差为d,则

an=a1+（n-1）d,

C^an=C^[a1+（n-1）d],

则C^an/C^a(n-1)=C^[a1+（n-1）d]/C^[a1+（n-2）d]=C^d 为定值,

而C^a1≠0,

所以{C^an}(c不等于0)为等比数列,公比为 q=C^d.