数列{an},{bn}中,an=lg(3^n)-lg[2^(n+1)],bn=a2n,求证{bn}是否为等差数列

题目

数列{an},{bn}中,an=lg(3^n)-lg[2^(n+1)],bn=a2n,求证{bn}是否为等差数列

答案

是等差数列
an=lg(3^n)-lg[2^(n+1)]
=nlg3-(n+1)lg2
bn-bn-1
=a2n-a2(n-1)
=2nlg3-(2n+1)lg2-{2(n-1)lg3-[2(n-1)+1]lg2}
=2nlg3-2nlg2-lg2-2nlg3+2lg3+2nlg2-lg2
=2lg3-2lg2为常数
所以根据等差数列的定义,bn为等差数列