F.4math !binomial theorem

1)(3+2x)^74 th term = t(3+1) = (7C3) [3^(7-3)] (2x)^3 = 22680x^35 th term = t(4+1) = (7C4) [3^(7-4)] (2x)^4 = 15120x^4T4:T5 = 22680 : 15120 = 3 : 2 2)(1-ax)(1-bx)^6For (1 - bx)^6 ,T(r+1) = (6Cr)(- bx)^r ,when r = 0 , T1 = (6C0)(- bx)^0 = 1when r = 1 , T2 = (6C1)(- bx)^1 = - 6bxwhen r = 2 , T3 = (6C2)(- bx)^2 = (15b^2)x^2So the coefficient of x = 1 * (- 6b) + (- a) * 1 = - a - 6b
the coefficient of x^2 = 1 * (15b^2) + (- a) * (- 6b) = 6ab + 15b^2- a - 6b = - 19.....(1)6ab + 15b^2 = 153.....(2)By (1) :a = 19 - 6b , sub it to (2) :6(19 - 6b)b + 15b^2 = 153114b - 36b^2 + 15b^2 = 15321b^2 - 114b + 153 = 07b^2 - 38b + 51 = 0(7b - 17)(b - 3) = 0b = 17/7 (rejected) or b = 3 ,a = 19 - 6*3 = 1

2010-08-26 22:40:23 補充：
Sorry , no need to reject b = 17/7 ,

when b = 17/7 ,

a = 19 - 6(17/7) = 31/7