The ninth term of the expansion left(3x - fra | vedclass

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(D) The general term of the expansion $(x+a)^{n}$ is $T_{r+1} = {}^{n}C_{r} x^{n-r} a^{r}$.Given the expansion $\left(3x - \frac{1}{2x}\right)^{8}$,we

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$\frac{1}{256x^{8}}$

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(D) The general term of the expansion $(x+a)^{n}$ is $T_{r+1} = {}^{n}C_{r} x^{n-r} a^{r}$.Given the expansion $\left(3x - \frac{1}{2x}\right)^{8}$,we have $n = 8$.To find the ninth term $(T_{9})$,we set $r+1 = 9$,which implies $r = 8$.Substituting the values $n = 8$,$r = 8$,$x = 3x$,and $a = -\frac{1}{2x}$ into the formula:$T_{9} = {}^{8}C_{8} (3x)^{8-8} \left(-\frac{1}{2x}\right)^{8}$$T_{9} = 1 \cdot (3x)^{0} \cdot \left(-\frac{1}{2x}\right)^{8}$$T_{9} = 1 \cdot 1 \cdot \frac{(-1)^{8}}{2^{8} x^{8}}$$T_{9} = \frac{1}{256x^{8}}$

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

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