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

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Grazes
 2 years ago
Best ResponseYou've already chosen the best response.0\[(\frac{ x ^{2} }{ 3 }  \frac{ 1 }{ x })^{9}\]

hewsmike
 2 years ago
Best ResponseYou've already chosen the best response.1Nasty one. The binomial expansion\[(a+b)^{n}= \sum_{k=0}^{n}\frac{ n! }{k!(nk)!}a^{nk}b^{k}\]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\[(a)^{nk}(b)^{k}=(x^{2}/3)^{9k}(x^{1})^{k}=constant \times x^{0}\]and so looking at those powers\[2(9k)+(k)=0\]\[182kk=0\]\[3k=18\]\[k=3\]so what is the multiplier of that power of x?\[(1/3)^{93}(1)^{3}\frac{9!}{3!(93)!}=(1)(3^{6})\frac{(9\times 8 \times 7 \times 6 \times 5\times4 \times 3 \times 2 \times 1)}{(3 \times 2 \times 1)(6 \times 5\times4 \times 3 \times 2 \times 1)}\].... aaargh ......\[=28/3^{5}\]I think ! :)

Grazes
 2 years ago
Best ResponseYou've already chosen the best response.0I just know that the final answer is 29/8. Thanks!

hewsmike
 2 years ago
Best ResponseYou've already chosen the best response.1Ah, I've probably fumbled the arithmetic somewhere, but the method is as described. :)
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