Question

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

Answer 1

See below:

Explanation:

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.