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)#