# Coefficient Of Expansion

1. Jul 31, 2008

### Air

1. The problem statement, all variables and given/known data
a) Find the coefficient of $x^5$ in $(1+x+x^2)^4$.

b) Find the coefficient of $x^2$ in $(2+2x+x^2)^n$.

2. The attempt at a solution
a) $(1)^0(x)^3(x^2)^1$ & $(1)^1(x)^1(x^2)^2$.
Coefficient of first: $\frac{4!}{(0!)(3!)(1!)} = 4$
Coefficient of second: $\frac{4!}{(1!)(1!)(2!)}=16$
Therefore coefficient of $x^5$ is $4+12 = 16$.

b) This is the part I'm struggling on. May I have help here. It's a general term (power of $n$ which is confusing me).

2. Jul 31, 2008

### tiny-tim

Hi Air!

(btw, "Coefficient Of Expansion" means something specific in materials physics)

Well, it's either x2 or x x;

if you use the same formulas as before, what do you get?

3. Jul 31, 2008

### Air

So would that be:

$(2)^0(2x)^0(x^2)^1$ and $(2)^1(2x)^2(x^2)^0$

What am I to do with the $(2)$? It can be used many times?

("Coefficient Of Expansion" was just a random title I thought of which might be suitable for this, had no idea that it's something in materials physics )

4. Jul 31, 2008

### tiny-tim

No, it's …

$$(2)^{n-1}(2x)^0(x^2)^1$$ and … ?

5. Jul 31, 2008

### Air

Hmmm.... I'm not sure how it's possible to get $(2x)^2$ and $(x^2)^0$ together as the powers have to be consecutive, don't they?

6. Jul 31, 2008

### tiny-tim

erm … I've no idea what you mean … but the answer is, no!

7. Jul 31, 2008

### Air

Would it be this:

$(2)^{n}(2x)^2(x^2)^0$

I'm sure about the $(2x)^2(x^2)^0$ because it would get $(4x^2)(1)$ which would give a coefficient of value $4$ but is the power of $2$ correct?

8. Jul 31, 2008

### tiny-tim

Hint: the exponents (that's the itsy-bitsy thingys) have to add to n, don't they?

9. Jul 31, 2008

### HallsofIvy

Staff Emeritus
Don't be sure- that term won't be in an nth power. In $(2+2x+x^2)^n$, every term is of the form Cijk(2)i(2x)j(x2)k with i+ j+ k= n and C is the "tri"-nomial coefficient:
$$C_{ijk}= \frac{n!}{i!j!k!}$$

One way to get x2 would be (2)n-2(2x)2(x2)0 but another would be (2n-1)(2x)0(x2)1. And don't forget the "tri"-nomial coefficient for each of those.

Last edited: Jul 31, 2008
10. Jul 31, 2008

### Air

Oh, I see. Thanks. That makes sense.

But I have another question:
As this question is $(2+2x+x^2)^n$, does that mean it's even possible to find the coefficient of $x^{235}$ (for instance) as $n$ can be any value?

11. Jul 31, 2008

### HallsofIvy

Staff Emeritus
Yes, it certainly should be. As I said before, Cijk(2)i(2x)j(x[sup2)k with i+ j+ k= n and C is the "tri"-nomial coefficient.

In order to get x235, you would have to have j+ 2k= 235. There are many values of n large enough to include that. Of course, with x2 any positive n is sufficient.