若f(x)=3x+4,f1(x)=f(x),f2(x)=f(f1(x)),f3(x)=f(f2(x)),···,···,fn(x)=f(fn-1(x)),n属于N*,则f(f(···f(x)))=?注:共n个
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![若f(x)=3x+4,f1(x)=f(x),f2(x)=f(f1(x)),f3(x)=f(f2(x)),···,···,fn(x)=f(fn-1(x)),n属于N*,则f(f(···f(x)))=?注:共n个](/uploads/image/z/5223417-33-7.jpg?t=%E8%8B%A5f%28x%29%3D3x%2B4%2Cf1%28x%29%3Df%28x%29%2Cf2%28x%29%3Df%28f1%28x%29%29%2Cf3%28x%29%3Df%28f2%28x%29%29%2C%C2%B7%C2%B7%C2%B7%2C%C2%B7%C2%B7%C2%B7%2Cfn%28x%29%3Df%28fn-1%28x%29%29%2Cn%E5%B1%9E%E4%BA%8EN%2A%2C%E5%88%99f%28f%28%C2%B7%C2%B7%C2%B7f%28x%29%29%29%3D%3F%E6%B3%A8%EF%BC%9A%E5%85%B1n%E4%B8%AA)
若f(x)=3x+4,f1(x)=f(x),f2(x)=f(f1(x)),f3(x)=f(f2(x)),···,···,fn(x)=f(fn-1(x)),n属于N*,则f(f(···f(x)))=?注:共n个
若f(x)=3x+4,f1(x)=f(x),f2(x)=f(f1(x)),f3(x)=f(f2(x)),···,···,fn(x)=f(fn-1(x)),n属于N*,则
f(f(···f(x)))=?注:共n个
若f(x)=3x+4,f1(x)=f(x),f2(x)=f(f1(x)),f3(x)=f(f2(x)),···,···,fn(x)=f(fn-1(x)),n属于N*,则f(f(···f(x)))=?注:共n个
(3^n)x+4+4*3+4*3^2+...+4*3^(n-1)
=(3^n)x + (4-4*3^(n-1))/(1-3)
=(3^n)x + 2*3^(n-1)-2