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

Click For Summary

Homework Help Overview

The problem involves finding the coefficient of x8 in the expansion of (x2 - 3)7. Participants are discussing the implications of the variables involved, particularly the confusion surrounding the values of n and j in the context of the binomial expansion.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to clarify the roles of n and j in the binomial expansion, questioning how n - j can yield a negative number. There is also discussion about the implications of using x2 instead of x, leading to confusion about the correct values for n and j.

Discussion Status

Some participants have provided guidance on interpreting the binomial expansion correctly, suggesting that the confusion arises from the use of the variable x in multiple contexts. Others have shared their attempts with different values for n and j, noting which combinations worked and why they believe certain values are appropriate.

Contextual Notes

There is an indication that the original problem may have been misinterpreted due to the notation used, particularly with the variable x. Participants are exploring the implications of this notation on their understanding of the problem.

rakeru
Messages
75
Reaction score
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!
 
Physics news on Phys.org
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.
 
  • Like
Likes   Reactions: 1 person
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! :)
 

Similar threads

Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
6
Views
10K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
Replies
15
Views
2K
Replies
34
Views
5K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K