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Math Help - Problem With Binomial Coefficients

  1. #1
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    Problem With Binomial Coefficients

    Hello,

    I am trying to solve the following problem:

    "Find the term independent of x in  (x+\frac{1}{x})^3)^8 ."

    I think this formula applies, but I am not sure where to begin:  t_{k+1}={n \choose k}x^{n-k}y^k.

    Can anyone help with this question?
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Problem With Binomial Coefficients

    Do you mean:

    \left(x+\left(\frac{1}{x} \right)^3 \right)^8 ?
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  3. #3
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    Re: Problem With Binomial Coefficients

    Quote Originally Posted by MarkFL2 View Post
    Do you mean:

    \left(x+\left(\frac{1}{x} \right)^3 \right)^8 ?
    Correct! Sorry about that!
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  4. #4
    MHF Contributor MarkFL's Avatar
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    Re: Problem With Binomial Coefficients

    Okay, the binomial theorem tells us the general term in this expansion will be:

    {8 \choose k}x^{8-k}\left(\left(\frac{1}{x} \right)^3 \right)^k={8 \choose k}x^{8-k}x^{-3k}={8 \choose k}x^{8-4k}

    Now, in order for this term to be independent of x, to what must the exponent on x be equal?
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  5. #5
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    Re: Problem With Binomial Coefficients

    Quote Originally Posted by MarkFL2 View Post
    Okay, the binomial theorem tells us the general term in this expansion will be:

    {8 \choose k}x^{8-k}\left(\left(\frac{1}{x} \right)^3 \right)^k={8 \choose k}x^{8-k}x^{-3k}={8 \choose k}x^{8-4k}

    Now, in order for this term to be independent of x, to what must the exponent on x be equal?
    0, correct? Therefore, would k be 2?
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  6. #6
    MHF Contributor MarkFL's Avatar
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    Re: Problem With Binomial Coefficients

    Yes, correct on both counts!

    So, what would the value of that term be?
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  7. #7
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    Re: Problem With Binomial Coefficients

    Quote Originally Posted by MarkFL2 View Post
    Yes, correct on both counts!

    So, what would the value of that term be?
    It would be {8 \choose 2} = 28?

    Thanks!
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  8. #8
    MHF Contributor MarkFL's Avatar
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    Re: Problem With Binomial Coefficients

    Yes, correct again!
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