求解二项式系数，（1+2x)(1-x)^4中x^2的系数。求详细过程。解释谢谢

Question

如题所述

Answer 1

（1+2x)(1-x)^4
=(1+2x)(1-x)²)(1-x)²
=（1+2x)(1-2x+x²)(1-2x+x²)
=(1-2x+x²+2x-4x²+2x^3)(1-2x+x²)
=(1-3x²+2x^3)(1-2x+x²)
=1-2x+x²-3x²+6x^3-3x^4+2x^3-4x^4+2x^5
=1-2x-2x²+8x^3-7x^4+2x^5
所以，（1+2x)(1-x)^4中x^2的系数为-2.

Answer 2

(1+2x)(1-x)^4
coef. of x^2
1. 4C2 - 2(4C1)
= 6-8
=-2

Answer 3

(1+2x)(1-x)^4中展开(1-x)^4=x^4-4x^3+6x^2-4x+1，这样只有6x^2和-4x有用，1*6x^2-2x*4x=-2x^2，所以x^2的系数为 -2