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[royal granite]

How do i go about solving this after expanding (a+b)^5

[weak walrus]

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[slim lodge]

notice when the coefficient of one of the terms is just x

wait no

notice when the value of x is just x

not x^2, or x^-2

[royal granite]

no i dont wheres that?

[slim lodge]

so think about this
what will the power of x be for a^3?

[royal granite]

im sorry im not following

[reef urchin]

First, what is the general form of the term in the expansion of (a + b)^5?

[royal granite]

a^{5} + 5a^{4}b + 10a^{3}b^{2} + 10a^{2}b^{3} + 5ab^{4} + b^{5} do u mean this?

[reef urchin]

That is the whole sum. I meant the term.

So, like the term of (x + y)^n would be a(k) = C(n, k) x^(n - k) y^k for k = 0, ..., n.

[royal granite]

so this T(r+1) = C(5, r) * a^(5-r) * b^r?

[reef urchin]

Right.
Notice how you can also rewrite it as C(5, r) a^5 (b/a)^r.

Ah, actually, nevermind. That way doesn't really help here.

Anyway, in your case a = 4x, b = 2/(9x).

So, substitute that into the expression of the term.

[royal granite]

so i substitute it back in to this?

[reef urchin]

No, into this.

[royal granite]

oh

so it becomes this?

T(r+1) = C(5, r) * 4^(5-r) * 2^r * (1/(9^r)) * x^5.

[reef urchin]

No, I don't think that's it. Let me try, though.

a(k) = C(5, k) (4x)^(5 - k) (2/(9x))^k = C(5, k) 4^(5 - k) (2/9)^k x^(5 - k - k) = 4^5 C(5, k) (1/8)^k x^(5 - 2k)

[royal granite]

can u show me ur wroking out to get there?

[reef urchin]

I did, though.

[royal granite]

yea i realised after writing it down and looking at it

but is that all to the question or is there more?

[reef urchin]

Well, of course that's not it.

We are asked to find the coefficient of x.

So, x^1.

Note that the term a(k) has x^(5 - 2k).

So, you can find k - the index of this term.

[royal granite]

i cant seem to work it out

[reef urchin]

Well, you just need to solve 5 - 2k = 1...

[royal granite]

so k = 2?

[reef urchin]

Yes.

[royal granite]

+close