数列满足a1=1,a2=2,a(n+2)=(1+cos)数列{an}满足a1=1,a2=2,a(n+2)=(1+cos(nπ/2)^2)an+sin(nπ/2)^2,n=1,2,3,……求a3,a4,并求数列{an}的通项公式.

Question

数列满足a1=1,a2=2,a(n+2)=(1+cos)
数列{an}满足a1=1,a2=2,a(n+2)=(1+cos(nπ/2)^2)an+sin(nπ/2)^2,n=1,2,3,……
求a3,a4,并求数列{an}的通项公式.

Answer 1

题目条件应该是a(n+2)={1+[cos(nπ/2)]^2}an+[sin(nπ/2)]^2
如果是的话,答案如下：
易求a3=2,a4=4
由题设a(n+2)=[1+(cosnπ+1)/2]an+(1-cosnπ)/2
∴ n=2k (k∈N+)时：a(n+2)=2an
n=2k+1 (k∈N+)时：a(n+2)=an+1
∴ 2^(n/2) n∈N+
an=｛
(n+1)/2 n∈N+