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Here is a review problem I am currently working on for my upcoming exam:

Show that for $m = 4n, 4n-3,4n-6, ... , -2n$, the coefficient of $x^m$ in ($x^2+\frac{1}{x})^{2n}$ is $$ \frac {(2n)!}{\left(\frac{4n-m}{3}\right)!\left(\frac{2n+m}{3}\right)!}$$

I've been scratching my head for quite some time on this problem. I've tried expanding ($x^2+\frac{1}{x})^{2n}$ as a binomial series but I still haven't been able to do anything effective at this point.

EDIT: There are suppose to be parenthesis around the fractions ($\frac{4n-m}{3}$) and ($\frac {2n+m}{3}$), but for some reason, it's not reading it correctly. Perhaps I didn't place them correctly.

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  • $\begingroup$ Hint: your fraction is $\binom{2n}{\frac{m+2n}{3}}$ (i.e. $_{2n}C_{\frac{m+2n}{3}}$) $\endgroup$ – Mosquite Feb 22 '17 at 0:09
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Expanding the binomial series does help.

$$(x^2+x^{-1})^{2n} = \sum_{k=0}^{2n} \dfrac{(2n)!}{k!\,(2n-k)!} x^{3k-2n}$$

Now you need $m=3k-2n$ solved for $k$ for each of the given $m\in\{4n-3i: i\in\{0,..\}\}$

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  • $\begingroup$ Ahh seeing that summation really brightens my day. $\endgroup$ – John Smith Feb 22 '17 at 0:33
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Write your $(x^2+\frac{1}{x})^{2n}$ as $\frac{1}{x^{2n}}(x^3+1)^{2n}$. Using binomial theorem this equals $$\binom{2n}{0}\frac{x^{6n}}{x^{2n}}+\binom{2n}{1}\frac{x^{6n-3}}{x^{2n}}+\binom{2n}{2}\frac{x^{6n-6}}{x^{2n}}+\ldots+\binom{2n}{2n}\frac{x^{0}}{x^{2n}}$$ That is, each term can be written as $$\binom{2n}{i}\frac{x^{6n-3i}}{x^{2n}}$$ for $i\in\{0,\ldots,2n\}$. The rest is algebra using $x^{m}=\frac{x^{6n-3i}}{x^{2n}}$, or, equivalently, $m=4n-3i$ and hence $i=\frac{4n-m}{3}$.

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  • $\begingroup$ Im not understanding how you got from $(x^2+\frac{1}{x})^{2n}$ to $\frac{1}{x^{2n}}(x^3+1)^{2n}$. $\endgroup$ – John Smith Feb 22 '17 at 0:35
  • $\begingroup$ @i8Σπ_821 Since $x^2+\frac{1}{x}=\frac{1}{x}(x^3+1)$. $\endgroup$ – Jan Feb 22 '17 at 0:39
  • $\begingroup$ Ahh okay now I see it. $\endgroup$ – John Smith Feb 22 '17 at 0:46

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