设函数f(x)=ax+b,其中a,b是实数,f1(x)=f(x),fn+1(x)=f(fn(x)),n=1,2,3,…若f7(x)=128x+381,则a+b=_.

问题描述:

设函数f(x)=ax+b,其中a,b是实数,f1(x)=f(x),fn+1(x)=f(fn(x)),n=1,2,3,…若f7(x)=128x+381,则a+b=______.

∵f(x)=ax+b,∴f2(x)=f[f(x)]=a(ax+b)+b=a2x+b(a+1),f3(x)=f[f2(x)]=a[a2x+b(a+1)]+b=a3x+ab(a+1)+b=a3x+b(a2+a+1)f4(x)=f[f3(x)]=a[a3x+b(a2+a+1)]+b=a4x+b(a3+a2+a+1)…f7(x)=a7x...