答：[(-1)^(n+1)]/[n(n+1)]方法一记f(x)=(x-1)g(x),其中g(x)=[(x-2)...(x-n)]/[(x+1)(x+2)...(x+n)],当n为奇数时g(1)=(n-1)!/(n+1)!=1/[n(n+1)],当n为偶数时g(1)=-(n-1)!/(n+1)!=-1/[n(n+1)],求导f'(x)=g(x)+(x-1)g'...