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

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SUMMARY

The discussion focuses on finding the coefficient of x8 in the expansion of (x2 - 3)7. The correct approach involves using the binomial theorem, specifically the formula (a + b)n = Σ (n choose j) aj b(n-j). Participants clarify that x should be treated as x2 and that n remains 7 while j must be 4 to avoid negative powers. The final coefficient is derived correctly by substituting these values into the binomial expansion.

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  • Understanding of the binomial theorem
  • Familiarity with binomial coefficients (n choose j)
  • Basic algebraic manipulation of exponents
  • Experience with polynomial expansions
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  • Study the binomial theorem in detail, focusing on its applications in polynomial expansions.
  • Learn about binomial coefficients and how to calculate (n choose j).
  • Practice problems involving coefficients in polynomial expansions, particularly with squared terms.
  • Explore the use of computational tools like Wolfram Alpha for verifying mathematical solutions.
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Students studying algebra, particularly those tackling polynomial expansions and the binomial theorem, as well as educators looking for clarification on teaching these concepts.

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

Thank you!

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

Thanks! :)
 

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