已知函数f(x)=3(x-1)/2,若数列an满足a(n+1)=f(an)·a1=2 (1)求an (2)若bn=(an)^2+2,求数列{bn}的最小项

题目

已知函数f(x)=3(x-1)/2,若数列an满足a(n+1)=f(an)·a1=2 (1)求an (2)若bn=(an)^2+2,求数列{bn}的最小项
已知函数f(x)=3(x-1)/2,若数列an满足a(n+1)=f(an)·a1=2
(1)求an (2)若bn=(an)^2+2,求数列{bn}的最小项
哪位高手帮下啊.头大

答案

(1)f(x)=3(x-1)/2,a(n+1)=f(an),所以a(n+1)= 3(an -1)/2,则a(n+1)-3=3/2*(an-3),数列{ an-3}构成等比数列,首项为a1-3=-1,公比是3/2.所以an-3= -(3/2)^(n-1)an= -(3/2)^(n-1)+3.(2)bn=(an)^2+2,设(3/2)^(n-1)=t,bn=(-...