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Math Help - Binomial expansion

  1. #1
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    Binomial expansion

    Could someone give me guidelines on doing this question?
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  2. #2
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    Do you understand that (1+x)S=S+xS?
    S=(1-x)^9= \sum\limits_{j = 0}^9 {\binom{9}{j}\left( { - 1} \right)^j x^j } .
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  3. #3
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    That makes sense but I can't get the answer.
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  4. #4
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    What can't you do? Post something of your own.
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  5. #5
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    I've probably not added the two Sigmas properly.
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  6. #6
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    Quote Originally Posted by Stuck Man View Post
    I've probably not added the two Sigmas properly.
    The coefficient of x^r is:


    \displaystyle {9 \choose r} (-1)^r + {9 \choose r - 1} (-1)^{r-1}.


    Your job is to simplify this and hence get the answer.

    Note:

    \displaystyle \frac{9!}{r! (9 - r)!} =  \frac{9! (10 - r)}{r! (10 - r)!}

    \displaystyle \frac{9!}{(r-1)! (10 - r)!} =  \frac{9! r}{r! (10 - r)!}
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  7. #7
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    I have found the answer except I have 9-r instead of 10-2r.
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  8. #8
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    I have solved it now. Thanks.
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  9. #9
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    Is there anything wrong with this?

    I don't understand why when I substitute x for 3 I get -564950498 and when I use the books answer with my calculator I get -524596891. When I substitute x for 2 the results are identical.
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  10. #10
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    Sorry I'm talking about a different question now. I've not given the books answer for it. Anyway is there an accuracy problem with my calculator?
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  11. #11
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    I have not included x^11 in the summation.
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  12. #12
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    The books answer is wrong. Here is the correct answer and working.
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  13. #13
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    I've shown that there can be an expression involving x to the power r in a sigma expression. A term containg x to the power r and not any other power of x can not be found for the rth element. The books question doesn't make sense. The book has many errors in the answers section for this group.
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  14. #14
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    It can actually be done.
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  15. #15
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    The book does have almost the correct answer for the first question. It should be sigma from r=0 to 10.
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