Given that, the coefficient of x^4 is 3840 in the binomial expansion of (1+qx)^6. Find:

Question

i. The value of q.
ii show that the of sum of kx^3 and px^4 is 5120.

Answer 1

The elements of row 6 of Pascal's triangle are 1,6,15,20,15,6,1
So the x^4 term is 15(qx)^4 = 15q^4 x^4. So,
15q^4 = 3840
q = 4

No idea what k and p have to do with it, but
20*4^3 + 15*4^4 = 5120

Answer 2

I think you are looking at the actual expansion of (1+4x)^6 , which we now know to be:

1 + 6(4x) + 15(4x)^2 + 20(4x)^3 + 15(4x)^4 + ....
= 1 + 24x + 240x^2 + 1280x^3 + 3840x^4 + ... thus confirming ooblecks first answer

so kx^3 and px^4 = 1280x^3 and 3840x^4
and the sum of their coefficients is 1280+3840 = 5120 , as needed

(of course you could not add the actual terms)