原函数fn(x)=x^n-(x-a)^n,求导得fn'(x)=n[x^(n-1)-(x-a)^(n-1)]=nfn-1(x)则fn+1'(x)=(n+1)fn(x)所以fn+1'(n+1)=(n+1)fn(n+1)因为f1(x)=a＞0,且n为正整数,所以fn'(x)＞0,即fn(x)是单调增函数所以fn(n+1)＞fn(n)故,fn+...