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

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rakeru
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Homework Statement



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

Homework Equations



This one:

(n n-j) an-jxj



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.

Please help!
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!
 
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rakeru said:

Homework Statement



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

Homework Equations



This one:

(n n-j) an-jxj



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.

Please help!
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!

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.
 
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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.
 
In that case, j would be just four, right?
 
Oh my god! Yes! I see!

Thank you!

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

Thanks! :)