1.数列{An}中A1=2,A2=-1,An^2=A(n+1)*A(n-1)(n≥2),则An=?2.数列{an},{bn}中,已_数学解答

1.数列{An}中A1=2,A2=-1,An^2=A(n+1)*A(n-1)(n≥2),则An=?
2.数列{an},{bn}中,已知an=lg(3^n)-lg2^(n+1),bn=a3n,试问数列{bn}是等差数列吗?如果不是,请说明理由

人气：304 ℃　时间：2020-05-22 05:01:27

解答

(1) An^2=A(n+1)*A(n-1);
An/A(n-1)=A(n+1)/An; 所以,
An=A1*(A2/A1)*(A3/A2)*……*(An/A(n-1))
=A1*(A2/A1)^(n-1)=2*(-1/2)^(n-1)
(2)
bn-b(n-1)=a(3n)-a(3n-3)
=lg(3^(3n))-lg(2^(3n+1))-lg(3^(3n-3))+lg(2^(3n-2))
=3n*lg3-(3n+1)*lg2-(3n-3)*lg3+(3n-2)*lg2
=3lg3-3lg2=常数
所以{bn}是等差数列.