已知数列{an}满足:a1=1,a2=2,且 an+2=(2+cosnπ)(an-1)+3,n∈N+

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已知数列{an}满足:a1=1,a2=2,且 an+2=(2+cosnπ)(an-1)+3,n∈N+
求通项公式an

a(n+2)=(2+cosnπ)(an -1)+3
if n is odd
a(n+2)=(2+cosnπ)(an -1)+3
a(n+2)=(2-1)(an -1)+3
= an +2
a(n+2) -an = 2
an - a1= n-1
an = n
if n is even
a(n+2)=(2+cosnπ)(an -1)+3
= 3(an -1)+3
a(n+2) = 3an
an/a2 = 3^[(n-2)/2]
an = 2.3^[(n-2)/2]
ie
an = n ,if n is odd
=2.3^[(n-2)/2] ,if n is even
再问：翻译为我看的懂得好么
再答： a(n+2)=(2+cosnπ)(an -1)+3
若 n 是奇数
a(n+2)=(2+cosnπ)(an -1)+3
a(n+2)=(2-1)(an -1)+3
= an +2
a(n+2) -an = 2
an - a1= n-1
an = n

若 n 是偶数
a(n+2)=(2+cosnπ)(an -1)+3
= 3(an -1)+3
a(n+2) = 3an
an/a2 = 3^[(n-2)/2]
an = 2. 3^[(n-2)/2]
ie
an = n ,若 n 是奇数
=2. 3^[(n-2)/2] ,若 n 是偶数

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