Binomial Expansion Question Driving Me Mad!!!

[Acolyte]

This is the last question on my maths sheet, and i must have been staring at it for hours, ive read all my notes and book but i just cant piece it together at all.

Driving me crazy!

You guys got any pointers, im really confused.

 

[KNiVES]

The answer is 42.

 

[Acolyte]

Lol cheers mate.

 

[Vixen]

It's been a few years since I did stuff like this, but I might be able to help a little. How much do you understand? Do you know where to start from?
It's not a great help, but the first formula that you need is on wiki:
http://en.wikipedia.org/wiki/Binomial_theorem
Going through that equation, you need the term where k=0.

 

[DaveF]

For part (a), you want to expand out (1+x)^n (1+x)^n and find out what the coefficient of x^n is.

This is really the only difficult bit, once you've done this, (b) and (c) are easy.

 

[Acolyte]

It's been a few years since I did stuff like this, but I might be able to help a little. How much do you understand? Do you know where to start from?
It's not a great help, but the first formula that you need is on wiki:
http://en.wikipedia.org/wiki/Binomial_theorem
Going through that equation, you need the term where k=0.

Hmmm i expanded it like this.

(1+x)^n = 1 + nx + n(n-1)(x^2)/2 +...+x^n

For part (a), you want to expand out (1+x)^n (1+x)^n and find out what the coefficient of x^n is.

This is really the only difficult bit, once you've done this, (b) and (c) are easy.

The first bit is the bit im struggling with.

 

[Visage]

Hmmm i expanded it like this.

(1+x)^n = 1 + nx + n(n-1)(x^2)/2 +...+x^n

Yup, so if that expression is squared, then the x^n term will be the sum of the products of the x^y and x^(n-y) terms for 0<y<n.

 

[kitten_caboodle]

*head explodes*

 

[Acolyte]

Yup, so if that expression is squared, then the x^n term will be the sum of the products of the x^y and x^(n-y) terms for 0<y<n.

You lost me here, where has the y come from?

 

[Vixen]

You lost me here, where has the y come from?

I think he is using y in the same way k is used in the equation I linked.

EDIT: yeah, I've just had a look over it and that would be a normal way of expressing it if you were using the format delta(k=0 to n).

 

[Visage]

You lost me here, where has the y come from?

The y is the summation variable.

(1+x)^2n = (1+x)^n * (1+x)^n, right?

Now, we know that:

(1+x)^n = Sigma (k=0 to k=n) (n,k)x^k

Where (n,k) is the binomial co-efficient. With me still?

so, what you need to do is imagine multiplying the two infinite series together, and collecting coefficients.

Its like if you get (a+b+c) * (a+b+c), you get aa + ab + ac + ba + bb + bc + ca +cb +cc.

Same here, but you're doing it with a lot more terms.

Its one of those things thats really hard to explain, but quite easy to write down. I've just solved the problem (part 1, anyway), and its about 4 lines. I wont post it though, as copying it would teach you nothing.

 

[DaveF]

Hmmm i expanded it like this.

(1+x)^n = 1 + nx + n(n-1)(x^2)/2 +...+x^n

That's making life harder for yourself than you need to. Keep it in terms of the binomial coefficients nCk = n!/((n-k)!k!).

(1+x)^n = 1 + nC1 x + nC2 x^2 + ... + nCn = sum_{k=0 to n} nCk x^k

Now just write the whole thing out twice, bearing in mind you'll want to use a different index to sum over instead of 'k' for one of them; let's use 'j' instead.

Then (1+x)^n (1+x)^n = [sum_{j = 0 to n} nCj x^j] [sum_{k=0 to n} nCk x^k].

Now collect up the terms in x^n (hint if x^j x^k = x^n, what's k in terms of j).

Not sure how to go any further without giving the entire game away...