In the binomial expansion of (x/3 – 4/y)7, what is the coefficient of the term (x/y)4?

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Answer 1 · 2018-05-06T09:23:38

See below:

Let's do the expansion and see what we have:

$(a+b)^n=(C_(n,0))a^nb^0+(C_(n,1))a^(n-1)b^1+...+(C_(n,n))a^0b^n$

$((7),(0))(x/3)^7(-4/y)^0=(1)(x^7/2187)(1)=(x^7)/(2187)$
$((7),(1))(x/3)^6(-4/y)^1=(7)(x^6/729)(-4/y^6)=-(28x^6)/(729y)$
$((7),(2))(x/3)^5(-4/y)^2=(21)(x^5/243)(16/y^2)=(112x^5)/(81y^2)$
$((7),(3))(x/3)^4(-4/y)^3=(35)(x^4/81)(-64/y^3)=-(2240x^4)/(81y^3)$
$((7),(4))(x/3)^3(-4/y)^4=(35)(x^3/27)(256/y^4)=(8960x^3)/(27y^4)$
$((7),(5))(x/3)^2(-4/y)^5=(21)(x^2/9)(-1024/y^5)=-(7168x^2)/(3y^5)$
$((7),(6))(x/3)^1(-4/y)^6=(7)(x/3)(4096/y^6)=(28672x)/(3y^6)$
$((7),(7))(x/3)^0(-4/y)^7=(1)(1)(-16384/y^7)=-16384/y^7$

Giving the entire expansion as:

$(x/3-4/y)^7=(x^7)/(2187)-(28x^6)/(729y)+(112x^5)/(81y^2)-(2240x^4)/(81y^3)+(8960x^3)/(27y^4)-(7168x^2)/(3y^5)+(28672x)/(3y^6)-16384/y^7$

There is no term that has a $(x/y)^4$ term within it. The closest we get are the $(2240x^4)/(81y^3)$ and $(8960x^3)/(27y^4)$ terms.