# Finding a Coefficient in an Expansion (n-j is negative?)

1. Oct 15, 2013

### rakeru

1. The problem statement, all variables and given/known data

Find the coefficient of x8 in the expansion of (x2-3)7

2. Relevant equations

This one:

(n n-j) an-jxj

3. The attempt at a solution
Hi!
Well, I know that
x= x2
a= -3
n= 7
and apparently, j=8.

This is what confuses me. n-j is a negative number.. how would that even work?? My guess is that the x2 has to do with it somehow... I'm thinking because the x is squared, then n wouldn't be 7.

I wrote this on wolfram alpha because I was like.. "is this possible or is it just a typo?" and it gave me an answer!

Thank you!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 15, 2013

### Dick

No, no. (a+b)^n=sum over j of (n j)a^j*b^(n-j). Put a=x^2 and b=(-3). You are likely getting confused because you are using the same symbol x for two different things. Saying x=x^2 is just silly.

3. Oct 15, 2013

### rakeru

Hmm well.. that's how my professor did it.

So after trying multiple times with different values for n and j, I got the right one. I ended up using n=7 and j=4. At first I tried n=14 and j=8. Nope. Then, I tried n=14 and j=4. Nope. Then, I tried n=7 and j=4! It worked... I'm still not so sure of why. I think the n is seven like normal.. but if I put 8 as j, then x will be to the power of 16. So I tried 4 and it worked..

Thanks.

4. Oct 15, 2013

### Office_Shredder

Staff Emeritus
rakeru, if I asked you to calculate the power of y4 in (y-3)7, do you see why this is the same as your original question?

5. Oct 15, 2013

### rakeru

In that case, j would be just four, right?

6. Oct 15, 2013

### Dick

Of course it would. Same for your original problem if you separate the two uses of x.

7. Oct 16, 2013

### rakeru

Oh my god! Yes! I see!!

Thank you!!!

I wonder why my teacher did it like that, though.

Thanks! :)