# Math Help - Binomial theporem : Finding the Coefficient

1. ## Binomial theporem : Finding the Coefficient

Me again.

Got stuck on my last practice question and its because I have literally no idea what it is asking me.

The question says:

Find the coefficient of $x^9$in the expansion of $(x-\frac{2}{x})^1^5$

I Literally have no clue. If someone could try and talk me through this id really appreciate it.

2. The binomial theorem states that

$\displaystyle (a+b)^n=\sum_{k=0}^{n}\binom{n}{k}a^{k}b^{n-k}$

So in your case you want

$\displaystyle (x)^{k}\left( \frac{2}{x}\right)^{n-k}=(\text{some number})x^{9}$

Since we know that $n=15$ use this to solve for k and then find the coeffient.

3. You want to solve this $\dbinom{15}{k}\left(x^k\right) \left(\frac{-2}{x}\right)^{15-k}=\mathcal{C}x^9$.
Find $k~\&~\mathcal{C}$.
The answer is $\mathcal{C}$

4. Sorry guys...

At the risk of sounding thick: I really don't have a clue where to begin.

5. Originally Posted by Tauron
At the risk of sounding thick: I really don't have a clue where to begin.
If $x^k \left( {x^{k - 15} } \right) = x^9$ then what is $k~?$

6. 12?

7. Good for you!
Now $\mathcal{C}=\dbinom{15}{12}(-2)^{15-12}$

8. would the (15/12) be 2520?

If I remember its something to do with the factorials and cancelling. Not sure if Im on the right lines though.

9. I mean you no disrespect.
But you have absolutely no business doing this question if you do not know that $\dbinom{15}{12}=\dfrac{15!}{12!\cdot 3!}$.

10. Dont worry about it Plato! If it were my choice I wouldnt be doing these types of questions.

Perhaps a bit of background might help. Im having to do a Maths course at University as it is a required credit to pass out this year. However, maths was not required to be on my course (English Lit!) and at no point have I ever sat a decent level of math before.

Now if you could bear with me in the understanding that I was only introduced to the concept of a "factorial" and a "binomial expansion" in my last lecture then that would be appreciated!