Find the coefficient of x^12 in (1/x- 3x^2)^24

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The discussion revolves around solving three mathematical problems, including finding the coefficient of x^12 in the expression (1/x - 3x^2)^24. Participants offer guidance on using the binomial theorem and clarify concepts related to subsets and their sizes. The original poster expresses initial confusion but ultimately reports successfully solving the problems independently. Key points include understanding the fraction of subsets containing a specific element and correctly interpreting terms in polynomial expansions. The conversation highlights the importance of determination and seeking help when needed.
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hi, i need some help with the following questions! I would greatly appreciate it and the sooner the better! I really don't know how to do these, I'm basically stuck on them.. I have tried to solve them but I don't get the correct answer. Thank you so much! :smile:

1) a universal set of length 'n' contains the letter 'A'. Prove that the fraction of the subsets of size 'r' that include the letter 'A' is (r/n).

2)in the expansion of (2x+1/x), find the term that does not include x.

3) Find the coefficient of x^12 in (1/x- 3x^2)^24
 
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This looks like a HW problem to me...What are your ideas for each of the three problems...?Post your work.

Daniel.
 
So what have you tried? What thoughts did you have?
 
well for #3 the main problem i had was that it did not have an exponent and when i went in for help to see a math teacher at my school, she told me that it was "not possible" to do the question since there is no exponent. This question I had to use the binomial theorem and i got the answer for #3 to be 1, meaning the second term, but then the answer seems like it could also be 0, meaning the first term.

for the first question I'm stuck because I'm not sure of how to go about starting the proof, like what should i start with and what should be my initial assumptions.

well thank you for taking the time to look at my questions! :smile:
 
oops sorry i meant that for #2 it did not have an exponent! sorry sorry!
 
1. How many subsets does a set with n elements have? (You should have a formula for that)? How many of those do NOT contain "A"? (If you remove "A" from the set you now have n-1 elements. How many subsets does a set with n-1 elements have?)
Now, how many subsets DO contain "A" (answer to the first question minus answer to second question)? What fraction is that of all subsets.

2. If a term "does not include x", then you can think of it as x0.

3. If you know the binomial theorem, this should be easy.
 
Mika-Yugo said:
oops sorry i meant that for #2 it did not have an exponent! sorry sorry!

So you mean it is simply
2x+\frac{1}{x}

It doesn't make any sense... :bugeye:

Daniel.

P.S.Check the text again...
 
HallsofIvy said:
1. How many subsets does a set with n elements have? (You should have a formula for that)? How many of those do NOT contain "A"? (If you remove "A" from the set you now have n-1 elements. How many subsets does a set with n-1 elements have?)
Now, how many subsets DO contain "A" (answer to the first question minus answer to second question)? What fraction is that of all subsets.

Halls, I think you misread the question. It asks about subsets of size r. The correct way should be :

a) What is the number of unrestricted subsets of size r from a set of size n ?
b) What is the number of unrestricted subsets of size r from a set of size (n-1) ?
c) Difference between a) and b), divided by a) would give the answer.

The expression should be :

\frac{{}^nC_r - {}^{n-1}C_r}{{}^nC_r} which can be simplified to get the required answer.
 
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Mika-Yugo said:
3) Find the coefficient of x^12 in (1/x- 3x^2)^24

For this part, as Halls said, use the Binomial theorem. Don't expand the whole thing out, just use the expression for the general term and solve for the coefficient of the term with x^12.
 
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Hey guys! Thanks so much, but guess what... i have managed to solve them myself, yes- determination pays off! lol- i guess sometimes it's worth to listen to my father-hahahaha
o well, thanks so much, and you all don't need to worry anymore about my questions... for now that is! lol
until my next posting... bye bye
 
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