Question

求(1+2x)^3 (1-x)^4的展开式中x^2的项的系数?

Answer 1

(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

Answer 2

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

Answer 3

(1+2x)^3常数项是1^3=1，x项系数是3×2^1=6，x^2项系数是3×2^2=12x，
(1-x)^4后者常数项是1^4=1，x系数是4×(-1)^1=-4，x^2系数是6×(-1)^2=6
所以系数是12-24+6=-6

Answer 4

-6

Answer 5

-6