设f(x)=a0xn+a1xn-1+…+an-1x+an,则f(n)(0)____．
n!a0。解析：f′(x)=na0xn-1+(n-1)a1xn-2+．…．+an-1，f(n)=n-(n-1)•…•1•a0xn-n=n!a0，所以f(n)(0)=n!a0．
设f(x)=a0xn+a1xn-1+…+an-1x+an,则f(n)(0)____．
n!a0。解析：f′(x)=na0xn-1+(n-1)a1xn-2+．…．+an-1，f(n)=n-(n-1)•…•1•a0xn-n=n!a0，所以f(n)(0)=n!a0．
