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

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The discussion focuses on solving the binomial expansion of the expression (2x - 1/(2x^2))^9. The formula for the general term T_{r+1} is provided, which involves binomial coefficients and powers of the variables. The term is expressed as T_{r+1} = (9!/(r+1)!(8-r)!) * (-1)^{r+1} * 2^{7-2r} * x^{6-3r}. A key point is the relationship established between r and h, leading to the equation 3r - h = 9. The discussion seeks assistance in further simplifying or solving this binomial expansion problem.
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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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