Please help for binomial expansion (2x-1/(2x^2))^9

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The discussion focuses on the binomial expansion of the expression \((2x - \frac{1}{2x^2})^9\). The formula for the general term \(T_{r+1}\) is derived as \(T_{r+1} = \binom{9}{r+1} (2x)^{8-r} \left(-\frac{1}{2x^2}\right)^{r+1}\), simplifying to \(T_{r+1} = \frac{9!}{(r+1)!(8-r)!} (-1)^{r+1} \cdot 2^{7-2r} x^{6-3r}\). The discussion also highlights the relationship between the variables \(r\) and \(h\) through the equation \(3r - h = 9\).

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As titled, been cracking my head over it.
Thanks in advance!

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Here’s the first part of the question to get you going ...

For $(a-b)^9$, $\displaystyle T_{r+1} = \binom{9}{r+1} a^{9-(r+1)}(-b)^{r+1}$

$T_{r+1} = \dfrac{9!}{(r+1)!(8-r)!} (2x)^{8-r} \left(-\dfrac{1}{2x^2}\right)^{r+1} = \dfrac{9!}{(r+1)!(8-r)!} (-1)^{r+1} \cdot 2^{7-2r}x^{6-3r}$

Multiplying this term by $x^h$ would yield the variable factor as $x^{6-3r+h} = x^{-3} \implies 3r-h=9$
 

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