f(x)=x^2(x-x^3/3!+x^5/5!-x^7/7!+...+(-1)^k*x^(2k+1)!/k!+...) (k=0,1,...)
=x^3-x^5/3!+x^7/5!-x^9/7!+...+(-1)^k*x^(2k+3)/k!+...(k=0,1,...)
所以f^(n)(0)= 0 n为偶数或1; (-1)^k/k!n=2k+3 (k=0,1,...)