How do you find the coefficient of #a# of the term #ax^2y^8# in the expansion of the binomial #(4x-y)^10#?

1 Answer
Mar 5, 2017

Answer:

Coefficient #a# in the term #ax^8y^2# is #720#.

Explanation:

The terms of a binomial expansion of #(x+a)^n# are given by the series (with #(n+1)# terms#

#Sigma_(r=0)^(r=n)((n),(r))x^(n-r)a^r#, where #((n),(r))=(n!)/((n-r)!r!)#

Hence expansion of #(4x-y)^10# is given by

#Sigma_(r=0)^(r=10)((10),(r))(4x)^(10-r)(-y)^r#

As we are coefficient of #a# of the term #ax^2y^8#

#10-r=2# or #r=8# and it will occur in the #(8+1)^(th)# term of the sequence and this term will be

#((10),(8))(4x)^(10-8)(-y)^8#

= #(10!)/((10-8)!8!)(4x)^2(-y)^8#

= #(10xx9)/(1xx2)16x^2y^8#

= #720x^2y^8#

Hence, coefficient #a# in the term #ax^8y^2# is #720#.