a.math ( Binomial Theorem)

Yes. If n is not a positive integer, then you should reject the solution. This is because the binomial theorem in F4 additional mathematics holds for positive integral indices only.
SOLUTION
(1 + 2x)^n (1 - kx)^5
= [1 + n(2x) + nC2 * (2x)^2 + ...][1 - 5(kx) + 10(kx)^2 - ...]
= [1 + 2nx + 2n(n - 1) x^2 + ...][1 - 5kx + 10k^2 x^2 - ...]
= 1 + 2nx - 5kx + 2n(n - 1)x^2 - 10knx^2 + 10k^2 x^2 + ...
= 1 + (2n - 5k)x + [2n(n - 1) - 10kn + 10k^2]x^2 + ...
Therefore 2n - 5k = 0............(1)
2n(n - 1) - 10kn + 10k^2 = -60.......(2)
From (1), n = 5k / 2
Thus 2(5k / 2)(5k / 2 - 1) - 10k(5k / 2) + 10k^2 + 60 = 0
5k(5k - 2) / 2 - 25k^2 + 10k^2 + 60 = 0
5k(5k - 2) - 50k^2 + 20k^2 + 120 = 0
25k^2 - 10k - 50k^2 + 20k^2 + 120 = 0
-5k^2 - 10k + 120 = 0
-5(k^2 - 2k - 24) = 0
(k - 6)(k + 4) = 0
k = 6 or -4
When k = 6, n = 5(6) / 2 = 15.
When k = -4, n = 5(-4) / 2 = -10 (REJECTED as n is a positive integer)

2006-10-16 00:09:19 補充：
Yes. The factorization:
-5(k^2 +2k - 24) = 0
(k + 6)(k - 4) = 0
k = -6 or k = 4
When k = -6, n = 5(-6) / 2 = -15.(REJECTED as n is a positive integer)
When k = 4, n = 5(4) / 2 = 10

SORRY FOR ANY INCONVENIENCE CAUSED......