Question

已知数列{log2(a^n-1}为等差数列,且a1=3,a2=5。1。求证：数列{an-1}是等比数列。

Answer 1

解: 由{log2(a^n-1)}为等差数列
则设公差为d
则有: d=log2(a2-1)-log2(a1-1)
=2-1
=1
则有: log2(an-1)=log2(a1-1)+(n-1)d
=1+(n-1)
=n
则: an-1=2^n-------(1)
a(n-1)-1=2^(n-1)-------(2)
(1)/(2):
[an-1]/[a(n-1)-1]=2^n/[2^(n-1)]
=2
则数列{an-1}是首项为2,公比为2的等比数列.

Answer 2

数列{an-1}是等比数列。