已知数列《an》满足：a1＝1.A2＝2,且an+2＝（2+cosπ）（an-1）+3,n∈nж
问题描述：
已知数列《an》满足：a1＝1.A2＝2,且an+2＝（2+cosπ）（an-1）+3,n∈nж
求通项公式an
答
a(1)=1,a(2)=2,a(n+2) = [2 + cos(PI)][a(n)-1] + 3 = [2-1][a(n)-1] +3= a(n) + 2,a(2n+1) = a(2n-1+2) = a(2n-1) + 2,{a(2n-1)}是首项为a(1)=1,公差为2的等差数列.a(2n-1) = 1 + 2(n-1) = 2n-1.a(2n+2) = a(2n)+2,...