Tricky question from binomial expansion

Hi, I was doing a past year CIE paper and this question cropped up:

Find the value of k for which there is no term in x^2 in the expansion of (1+kx)(2-x)^6

Plz help!

Reply 1

gorilla_baby

Find the value of k for which there is no term in x^2 in the expansion of (1+kx)(2-x)^6
firstly, you expand the (2-x)^6 bracket
=2^6 + 6nCr1 x (2^5)x(-x) + 6nCr2 x (2^4)x(-x)^2
Then times it by the other bracker so (1+kx)(64 -192x +240x^2)
=64 -192x +240x^2 +64kx - 192kx^2 + 240kx^3
therefore, coefficient of x^2 will be (240-192k)x^2
240-192k=0
192k=240
k=1.25
plz tell me if thts the correct answer though..