標題:
coefficients 數學問題
發問:
此文章來自奇摩知識+如有不便請留言告知
given that the coefficients of x and x^2 in the expansion of (1-ax)(1bx)^6 are -19,153 find a and b 更新: 更正係（1-aX)(1-bx)^6 are -19,153
最佳解答:
(1 - ax)(1 - bx)6 = (1 - ax)(1 - 6bx + 15b2x2 - ...) Codfficient of x = - a - 6b = - 19 ... (1) Coefficient of x2 = 6ab + 15b2 = 153 ... (2) From (1): a = 19 - 6b Sub it into (2): 6b(19 - 6b) + 15b2 = 153 2b(19 - 6b) + 5b2 = 51 38b - 7b2 = 51 7b2 - 38b + 51 = 0 (7b - 17)(b - 3) = 0 b = 17/7 or 3 a = 13/7 or 1
其他解答:
Hope I can help you ! This question is used binomial theorem to answer : (1-ax)(1-bx)^6 = (1-ax)(1-bx+bx+.............) The required term of x = 1(-a)+1(-b) = -a-b = -19 The required term of x = (-a)(-b)+b = ab+b = 153 Therefore, we can get the following simultaneous equation -a-b = -19.................1 ab+b = 153..............2 From 2 , a = 19-b.......3 Substituting (3) into (2) b(19-b)+b = 153 19b = 153 b = 153/19...4 Substituting (4) into (1) -a-(153/19) = -19 a = 208/19 So , a = 208/19 b = 153/19|||||(1 ? bx)^6
留言列表
