A.Maths Binomial Theorem (show all steps)

(a)
(1-x)^11(1-5x^2)^2-(1+ax)^3(1+bx)^7
=(1-11x+55x^2-165x^3+...)(1-10x^2+25x^4)-(1+3ax+3a^2x^2+a^3x^3)(1+7bx+21b^2x^2+35b^3x^3+...)
=(1-10x^2-11x+110x^3+55x^2)-(1+7bx+21b^2x^2+35b^3x^3+3ax+21abx^2+63ab^2x^3+3a^2x^2+21a^2b^2x^3+a^3x^3)
=(1-11x-45x^2+110x^3)-(1+(7b+3a)x+(21b^2+21ab+3a^2)x^2+(35b^3+63ab^2+21a^2b^2+a^3)x^3)
=(7b+3a-11)x+(21b^2+21ab+3a^2-45)x^2+(35b^3+63ab^2+21a^2b^2+a^3+110)x^3
Since the coefficients of x andx^2 are zero
7b+3a-11=0...(1)
21b^2+21ab+3a^2-45=0...(2)
From (2)
7b^2+7ab+a^2-15=0
(1/7)(11-3a)(11-3a)+(11-3a)a+a^2-15=0
121-66a+9a^2+77a-21a^2+7a^2-105=0
5a^2-11a-16=0
(5a-16)(a+1)=0
a=16/5 or -1
b=1/5 or 2
(b)
Using 35b^3+63ab^2+21a^2b^2+a^3+110
when a=16/5, b=1/5
the coefficient of x^3 in the expansion
=159.7136
when a=16/5, b=1/5
the coefficient of x^3 in the expansion
=221





2007-10-17 22:16:13 補充：
最後是when a=-1, b=2the coefficient of x^3 in the expansion=221