Question

Please help me?

Answer 1

See below:

Explanation:

The binomial expansion has as its general form:

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

where the term `C_(n,k)` is a combination with general form:

`C_(n,k)=((n),(k))=(n!)/((k!)(n-k)!)` with `n="population", k="picks"`

So let's look at the expansion and see what it tells us.

As a first term, we have:

`x^7/128`

Notice that 128 can be written as `2^7`, so if we rewrite the expression, it's

`x^7/2^7`

If we put this into the first term of the expansion, we have:

`x^7/2^7=C_(n,0)a^nb^0=>C_(7,0)(x^7/2^7)(1)=>1(x/2)^7(1)`

So we now know that `n=7, a=x/2`

Let's now notice that the signs between the expanded terms are alternating plus to minus and back again. This means we're going to have a negative `b` term. And what is that term? We have:

`1/y^7=1^7/y^7`

Putting that into our general form, along with what we already know:

`1^7/y^7=C_(7,7)(x/2)^0(1^7/y^7)=>(1)(1)(1/y)^7`

And so `b=-1/y`

Altogether then, we have:

`(x/2-1/y)^7=x^7/128-(7x^6)/(64y)+(21x^5)/(32y^2)-...-1/y^7`

What have already is (aside from the two end terms above):

• `(C_(7,6))(x/2)^6(-1/y)^1=-(7x^6)/(64y)`
• `(C_(7,5))(x/2)^5(-1/y)^2=(21x^5)/(32y^2)`

And so the missing terms are:

• `(C_(7,4))(x/2)^4(-1/y)^3=-(35x^4)/(16y^3)`
• `(C_(7,3))(x/2)^3(-1/y)^4=(35x^3)/(8y^4)`
• `(C_(7,2))(x/2)^2(-1/y)^5=-(21x^2)/(4y^5)`
• `(C_(7,1))(x/2)^1(-1/y)^6=(7x)/(2y^6)`