Question

Find the $9th$ term in the expansion of ${(3x-\frac{1}{2x})}^8,x\ne 0.$

• A. $\frac{256}{x^8}$
• B. $\frac{\left(x^8\right)}{256}$
• C. $\frac{1}{256x^8}$
• D. $256x^8$

Answer 1

Answer: (C)

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

$T_9{=}^8{C}_8\left(3x{)}^{\circ }\right(\left(\frac{-1}{2x}{)}^8\right)=1\times 1\times \left(\frac{1}{2^8{x}^8}\right)=\left(\frac{1}{256x^8}\right)$