2條Binomial Theorem Question...(15分!!)今日完成...急急急!!

(1+ax)(1-2x)^6
=(1+3ax+3a^2 x^2+a^3x^3)(1-12x^2+...)
=1+3ax+(3a^2-12)x^2+(a^3-36a)^3+..

3a=k and (not enough information,please check the question clearly....)

2)

(1+2x)^n+(1-kx)^5
=[1+2nx+2n(n-1)x^2+...)(1-5kx+10k^2 x^2+...)
=1+(2n-5k)x+[10k^2-10nk+2n(n-1)]x^2+...

2n-5k=0 --(1)
and
10k^2-10nk+2n(n-1)=-60--(2)

by(1) , k=(2n)/5

sub k=(2n)/5 into (2) , we have

10(2n/5)^2-10(n)(2n/5)+2n(n-1)=-60
n^2+5n-150=0
(n-10)(n+15)=0
n=10 or n=-15 (rejected).

k=2n/5=20/5=4//

Question 1

(1+ax)3 (1-2x2)6 = (3C0 + 3C1a x + 3C2 a2x2 + 3C3 x3) (6C0 + 6C1 (-2x2) + 6C2 (-2x2)2 +…….+6C6 (-2x2)6)

= (1 + 3ax + 3 a2 x2 + a3 x3) ( 1 - 12x2 + 60x4 +…….+ 64 x 12)

= 1 + 3ax + 3 a2 x2 + a3 x3 - 12x2 - 36ax3 +……. (higher powers of x)

= 1 + 3ax + (3 a2-12) x2 + (a3 - 36a)x3 +……. (higher powers of x)

Given that the answer is = 1 + kx + term in x³ and other terms involving higher powers of x.

Therefore 3a = k and (3 a2-12) = 0 since the given coefficient of x2 is zero

By solving the equation, a = +2 or -2 and k = +6 or -6

substituting a and k given coefficient for x3 = a3 - 36 a = +64 or -64

The answer for question 2 was given by someone before