求(1+2x)^3 (1-x)^4的展开式中x^2的项的系数?

问题描述:

求(1+2x)^3 (1-x)^4的展开式中x^2的项的系数?

(1+2x)^3中x^2系数是C32×2^2=12,(1-x)^4常数项是1^4=1
(1+2x)^3中x系数是C31×2^1=6,(1-x)^4中x系数是C41×(-1)^1=-4
(1+2x)^3中x^2常数项是1^3=1,(1-x)^4系数是C42×(-1)^2=12
所以系数是12-24+12=0