# Binomial expansion

1. Dec 12, 2009

### icystrike

1. The problem statement, all variables and given/known data
Find the coefficient of $$x^{29}$$ in the expansion of $$(1+x^{5}+x^{7}+x^{9})$$.

2. Relevant equations

3. The attempt at a solution

2. Dec 12, 2009

### CompuChip

It is zero: there is no $x^{29}$ term in
$$(1+x^{5}+x^{7}+x^{9})$$

I suppose you are missing some power?

3. Dec 12, 2009

### icystrike

oh yea sorry.
its
$$(1+x^{5}+x^{7}+x^{9})^{16}$$

4. Dec 12, 2009

### CompuChip

So you have 16 factors of
$$(1+x^{5}+x^{7}+x^{9})$$
multiplying. The first thing I'd do is check which unique combinations of powers give 29. For example, suppose that you open up the brackets. You will encounter terms with four factors of x^5, a factor of x^9 and all 1's otherwise, which gives x^29. Are there any other combinations?

Then, go through them one by one... you are looking for something of the form
$$x^9 \cdot x^5 \cdot x^5 \cdot x^5 \cdot x^5 \cdot 1 \cdot 1 \ cdot 1 \cdots$$
(16 in total). How many different orders are there? I.e., when you again think about multiplying out the brackets, how many terms are there that give this expression?

5. Dec 12, 2009

### icystrike

I tried using another method:

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6. Dec 12, 2009

### CompuChip

Yeah, well, that was essentially what I was thinking of too.

7. Dec 12, 2009

### icystrike

thanks compuchip (=