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$The 9th term of the expansion of {(3 x - \frac{1}{2 x})}^{8} is$

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1256x8

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Correct option is D

3x-12x8 When we expand the given binomial, we get 9 terms, 9 th term is the last term. Tr+1=Cr n·xn-rar. for (x+a)r T9=C8 8(3x)0-12x8=+1256x8

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The 9th term of the expansion of 3x-12x8is

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$The 9th term of the expansion of {(3 x - \frac{1}{2 x})}^{8} is$