標題:
附加數問題
發問:
a)Find the values of the constants a and b if, in the expansion of (1-x)^11(1-5x^2)^2-(1+ax)^3(1+bx)^7 in ascending powers of x, the coefficients of x and x^2 are zero. b)Evaluate the coefficient of x^3 in the expansion.
此文章來自奇摩知識+如有不便請留言告知
最佳解答:
(a) The expansion up to x^3 is (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=-1, b=2 the coefficient of x^3 in the expansion =221
其他解答:63D0B758E2D502CC
留言列表
