Question

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

Answer 1

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`.