Binomial Theorem Question 20點!

(1+2x)^n * (1-kx)^5

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

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

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

From the first equation, we have n=2.5k, substitute into the second equation

5k(2.5k-1)+10k^2-25k^2=-60

12.5k^2-5k-15k^2=-60

2.5k^2+50k-600=0

k^2+2k-24=0

(k+6)(k-4)=0

k=4 or k=-6 (rejected) and n=2.5k=10

題目怪怪的 係數沒有問題嗎?