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Grazes Group Title

Find the term in the binomial expansion of (((x^2)/3)-(1/x))^9 that is a constant.

  • one year ago
  • one year ago

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  1. Grazes Group Title
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    \[(\frac{ x ^{2} }{ 3 } - \frac{ 1 }{ x })^{9}\]

    • one year ago
  2. hewsmike Group Title
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    Nasty one. The binomial expansion\[(a+b)^{n}= \sum_{k=0}^{n}\frac{ n! }{k!(n-k)!}a^{n-k}b^{k}\]has powers of a decreasing while those of b increase as you go up in the index k. Consider for this problem that \[a=x^{2}/3\]and \[b= -1/x=-x^{-1}\]and \[n=9\]They are effectively asking for the value of k that gives\[(a)^{n-k}(b)^{k}=(x^{2}/3)^{9-k}(-x^{-1})^{k}=constant \times x^{0}\]and so looking at those powers\[2(9-k)+(-k)=0\]\[18-2k-k=0\]\[3k=18\]\[k=3\]so what is the multiplier of that power of x?\[(1/3)^{9-3}(-1)^{3}\frac{9!}{3!(9-3)!}=(-1)(3^{-6})\frac{(9\times 8 \times 7 \times 6 \times 5\times4 \times 3 \times 2 \times 1)}{(3 \times 2 \times 1)(6 \times 5\times4 \times 3 \times 2 \times 1)}\].... aaargh ......\[=-28/3^{5}\]I think ! :-)

    • one year ago
  3. Grazes Group Title
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    I just know that the final answer is 29/8. Thanks!

    • one year ago
  4. hewsmike Group Title
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    Ah, I've probably fumbled the arithmetic somewhere, but the method is as described. :-)

    • one year ago
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