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

In summary, the conversation discusses finding the coefficient of x8 in the expansion of (x2-3)7 and the confusion surrounding the use of symbols and powers in the equation. The conversation concludes with the realization that the same method can be applied to finding the coefficient of y4 in (y-3)7.
  • #1
rakeru
75
0

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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  • #2
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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  • #3
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
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
In that case, j would be just four, right?
 
  • #6
rakeru said:
In that case, j would be just four, right?

Of course it would. Same for your original problem if you separate the two uses of x.
 
  • #7
Oh my god! Yes! I see!

Thank you!

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

Thanks! :)
 

1. What is a coefficient in an expansion?

A coefficient in an expansion refers to the numerical value that is multiplied by a variable in an algebraic expression. It represents the scale or magnitude of the variable in the expression.

2. How do I find a coefficient in an expansion?

To find a coefficient in an expansion, you can use the binomial theorem or the polynomial expansion method. These methods involve identifying the term with the desired coefficient and simplifying the expression to isolate the coefficient.

3. What does n-j is negative mean in finding a coefficient in an expansion?

In the context of finding a coefficient in an expansion, n-j being negative means that the term being evaluated has a negative exponent. This indicates that the term is part of the denominator in a fraction and may require additional steps in the expansion process.

4. Why is it important to consider negative exponents when finding a coefficient in an expansion?

Negative exponents in an expansion indicate that the term is part of the denominator in a fraction. If these terms are not properly accounted for, the final result of the expansion may be incorrect. Therefore, it is important to carefully consider and handle negative exponents in the process of finding a coefficient in an expansion.

5. Can I use a calculator to find a coefficient in an expansion?

Yes, you can use a calculator to help with the calculations involved in finding a coefficient in an expansion. However, it is important to have a clear understanding of the underlying concepts and steps involved in the process to ensure accurate results.

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