（1-x)^5*(1+x+x^2)^4的展开式中x^7的系数为?

Question

（1-x)^5*(1+x+x^2)^4的展开式中x^7的系数为?

Answer 1

（1-x)^5*(1+x+x^2)^4
=(1-x)^4(1+x+x^2)^4*(1-x)
=[(1-x)(1+x+x^2)]^4*(1-x)
=(1-x^3)^4*(1-x)
=[(1-x^3)^2]^2*(1-x)
=(1-2x^3+x^6)^2*(1-x)
=(1+4x^6+x^12-4x^3+2x^6-4x^9)(1-x)
=(x^12-4x^9+6x^4-4x^3+1)(1-x)
=x^12-x^13-4x^9+4x^10+6x^4-6x^5-4x^3+4x^4+1-x
=-x^13+x^12+4x^10-4x^9-6x^5+6x^4+4x^4-4x^3-x+1,
所以（1-x)^5*(1+x+x^2)^4的展开式中x^7的系数为0.