(2009•虹口区一模)已知:f(x)=ax+b(a,b∈R),f1(x)=f(x),fn+1(x)=f[fn(x)](
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(2009•虹口区一模)已知:f(x)=ax+b(a,b∈R),f1(x)=f(x),fn+1(x)=f[fn(x)](n∈N*),若f5(x)=32x-93,则a+b=______.
![(2009•虹口区一模)已知:f(x)=ax+b(a,b∈R),f1(x)=f(x),fn+1(x)=f[fn(x)](](/uploads/image/z/16970078-38-8.jpg?t=%EF%BC%882009%E2%80%A2%E8%99%B9%E5%8F%A3%E5%8C%BA%E4%B8%80%E6%A8%A1%EF%BC%89%E5%B7%B2%E7%9F%A5%EF%BC%9Af%EF%BC%88x%EF%BC%89%3Dax%2Bb%EF%BC%88a%EF%BC%8Cb%E2%88%88R%EF%BC%89%EF%BC%8Cf1%EF%BC%88x%EF%BC%89%3Df%EF%BC%88x%EF%BC%89%EF%BC%8Cfn%2B1%EF%BC%88x%EF%BC%89%3Df%5Bfn%EF%BC%88x%EF%BC%89%5D%EF%BC%88)
由f1(x)=f(x)=ax+b,得到f2(x)=f(f1(x))=a(ax+b)+b=a2x+ab+b,
f3(x)=f(f2(x))=a[a(ax+b)+b]+b=a3x+a2b+ab+b,
同理f4(x)=f(f3(x))=a4x+a3b+a2b+ab+b,
则f5(x)=f(f4(x))=a5x+a4b+a3b+a2b+ab+b=32x-93,
即a5=32①,a4b+a3b+a2b+ab+b=-93②,
由①解得:a=2,把a=2代入②解得:b=-3,
则a+b=2-3=-1.
故答案为:-1
f3(x)=f(f2(x))=a[a(ax+b)+b]+b=a3x+a2b+ab+b,
同理f4(x)=f(f3(x))=a4x+a3b+a2b+ab+b,
则f5(x)=f(f4(x))=a5x+a4b+a3b+a2b+ab+b=32x-93,
即a5=32①,a4b+a3b+a2b+ab+b=-93②,
由①解得:a=2,把a=2代入②解得:b=-3,
则a+b=2-3=-1.
故答案为:-1
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