Find the quotient of terms containing x^28 and x^-4 in the expansion of (2x^6-3/x^2)^10?

1 Answer
Oct 1, 2017

103680 for x^28 and 1180980 for x^(-4)

Explanation:

Expansion of (a+b)^10 will have 11 terms and its r^(th) term is given by

C_r^10 a^r*b^(10-r)

Here a=2x^6 and b=-3/x^2

Hence r^(th) term is C_r^10 (2x^6)^r*(-3/x^2)^(10-r)

= C_r^10 2^r*x^(6r)*(-3)^(10-r)(x^(-2))^(10-r)

= C_r^10 2^r*x^(6r)*(-3)^(10-r)(x^(-20+2r))

= C_r^10 2^r*(-3)^(10-r)*x^(6r-20+2r)

If we have x^28. 6r-20+2r=28 i.e. 8r=48 or r=6 and corresponding term is

C_8^10 2^8*(-3)^2*x^28=(10xx9)/(1xx2)*256*9x^28=103680x^28

and if we have x^(-4). 6r-20+2r=-4 i.e. 8r=16 or r=2 and corresponding term is

C_2^10 2^2*(-3)^8*x^(-4)=(10xx9)/(1xx2)*4*6561x^(-4)=1180980x^(-4)