设f(x)=a0xn+a1xn-1+…+an-1x+an,则f(n)(0)____.
设f(x)=a0xn+a1xn-1+…+an-1x+an,则f(n)(0)____.
【正确答案】: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
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